Direct Current Circuit

Electric current

The electric current through an area is the rate of change of charge from one side of an area to the another. Its SI unit is Ampere (A).
If a charge \(\Delta\)q crosses an area in time \(\Delta\)t, the average electric current through the area is then,

Fig(1): Flow of charge through an conductor of circular cross-section.
\[I_{av}=\frac{\Delta q}{\Delta t}\] The instantaneous current at time t is, \[\begin{align*} I&=\text{lim}_{\Delta t \rightarrow 0} \frac{\Delta q}{\Delta t}\\ I&=\frac{dq}{dt}\\ \end{align*}\]

Current Density

The current density at any point in a conductor is defined as the current flowing per unit area perpendicular to the direction of flow. If I be the uniformly distributed current over an area A and is perpendicular to it, then the current density is,
\[J=\frac{I}{A}\] If normal vector (indicated in figure by \(\hat{n}\) )makes an angle \(\theta\) with the direction of the current, then the current density is,

Fig(2): Figure showing the normal to the area making an angle \(\theta\) with the direction of current.
\[J=\frac{I}{Acos\theta}\] From this, we can write \[I=\vec{J}.\vec{A}\] Current is a scalar quantity though it has both magnitude and direction. It is because it does not follow the rules of vector addition (Vector addition!). Current density is however a vector quantity.
The SI unit of current density is A/m².

Mechanism of metallic conduction

A metal contains large number of free electrons. The remaining material is a collection of relatively heavy positive ions called 'lattice'. When the electric field is not applied in a conductor (metals), the electron moves in a zig zag path. Due to the large number of free electrons moving in random directions, the number of electrons crossing an area from one side equals the number crossing from the other side in any given interval of time. The electric current through the area is thus 0.

When an electric field is applied inside the conductor, a force acts on each electrons in the direction opposite to the field. The electrons slowly drifts in the direction of the force. At each collision, the electron gains an additional velocity due to the electric field. As the time 't' between successive collision with the lattice is small, the electrons slowly and steadily drifts opposite to the applied field. If the electrons drifts a distance l in a long time t , the drift velocity is defined as, \[v_d=\frac{l}{t}\]

Relation between current and drift velocity : Consider a conductor of an area A and length l . Let n be the number of electrons per unit volume in a conductor.

Fig(3): Flow of electrons through the conductor of length \(l\) when battery is connected across its ends.
If V be the volume of the conductor then, \[V=Al\] The total number of electrons in V volume of the conductor is, \[\begin{align*} N&=nV\\ &=nAl\\ \end{align*}\] When a conductor is connected to the battery, electrons drifts in the conductor opposite to the direction of applied field with drift velocity v_d in time t .
So the electric current is, \[\begin{align*} I&=\frac{q}{t}\\ &=\frac{nAl}{t}\\ &=nv_d eA\\ \therefore I&=v_d enA\\ \end{align*}\] So, the relation between current and drift velocity is derived.
Now, the current density J is,
\[\begin{align*} J&=\frac{I}{A}\\ &=\frac{v_d enA}{A}\\ \therefore J&=v_d en\\ \end{align*}\]

Q.1 Discuss the mechanism of metallic conduction. Derive a relation betweem current and drift velocity of the electron. ( See the solution above. )

Ohm's law

It states that, 'the current through a conductor is always directly proportional to the potential difference applied to the device'. If I be the current flowing through the conductor and V be the potential difference across its ends, then \[\begin{align*} V &\propto I\\ V&=IR\\ \end{align*}\] Here, R is the constant of proportionality called the resistance of the conductor.
Physical meaning of Ohm's law : See the plot of I Vs V in the figure. Here, the ratio \(\frac{I}{V}\) gives the slope of straight line which is same for all value of V. Since \(R=\frac{I}{V}\), it means that resistance of the conductor is independent of the magnitude and polarity of the applied potential difference V .
Let's take a break for a thought here! What actually is resistance and what causes it?
Would you get the same current if you apply the same potential difference across the ends of two different materials? No. Though we use two materials with similar geometry, currents vary. It is due to the electrical resistance of the conductor, 'the characteristic of the conductor' due to which current in two different materials differ. Resistance is a property of an object.
\[R=\frac{V}{I}\] The SI unit of resistance is ohm (\(\Omega\)).
Resistance in a conductor is caused due to the collision of electrons with the lattice sites of the conductor. More the collision, more is the resistance of the conductor. Hence, the resistance depends on length and area of the conductor wire. A long wire offers more resistance than a shorter one as more collisions occur in longer wire. Also, a thick wire offers low resistance compared to thin wire as the thick wire provides more cross-section for the flow of electrons. So, the resistance of a conductor wire is directly proportional to its length and inversely proportional to its area.

Q.2 A copper wire has a diameter of 1.02 mm and carries a constant current of 1.67 A. If the density of free currents in copper is 8.5 \(\times\) 10²⁸ m³, calculate the current density and the drift velocity of the electrons. ( Click here for the solution )!

Resistivity / Specific resistance

Let l be the length of a conductor of resistance R , and A be its cross-sectional area. Then,
\[\begin{align*} R&\propto l \text{... (i)}\\ R&\propto \frac{1}{A} \text {... (ii)} \end{align*}\] Combining equation (i) and (ii), we get, \[\begin{align*} R&\propto\frac{l}{A}\\ R&=\frac{\rho l}{A} \text{... (iii)}\\ \end{align*}\] Here, \(\rho\) is the constant of proportionality called the resistivity of a material or specific resistance. From equation (iii), \[\rho=\frac{RA}{l}\] The unit of \(\rho\) is \(\Omega m\).
\(\rho\) only depends on the material of the conductor.
If A=1 m\(^2\) and l=1 m, \(\rho= R\).
Resistivity of a conductor is thus defined as the resistance of the conductor of unit cross-sectional area and unit length.

Q.3 Resistance of a wire of length 1 m, diamter 1 mm is 2.2 \(\Omega\). Calculate its resistivity and conductivity. ( Click here for the solution)!

Conductance

Conductance is defined as reciprocal of resistance. It is represented by C and is given by, \[C=\frac{1}{R}\] The SI unit of conductance is per ohm (\(\Omega^{-1}\).

Conductivity

Conductivity is defined as reciprocal of resistivity. It is denoted by \(\Sigma\) and is given by,\[\Sigma=\frac{1}{\rho}\]. The SI unit of conductivity is per ohm per meter (\(\Omega^{-1} m^{-1}\)).

Q.4 A tightly coiled spring having 75 coils, each 3.50 cm in diameter, is made of insulated metal wire 3.25 mm in diameter. An ohm meter connected across its opposite ends reads 1.74 \(\Omega\). What is the resistivity of the metal? Click here for the solution!

Temperature dependence of Resistance

Let \(R_1\) be the resistance of the conductor at \(\theta_1^{\circ}\) C and \(R_2\) be the resistance of the conductor at \(\theta_2^{\circ}\) C. Then, \[R_2=R_1(1+\alpha \Delta \theta\] Here, \[\alpha=\frac{R_2-R_1}{R_1 \Delta {\theta}} \] is called the temperature coefficient of resistance. Metals have positive temperature coefficient (i.e., resistance increases with the increase in temperature). Semiconductors have negative temperature coefficient of resistance (i.e., resistance decreases with increase in temperature and vice-versa).

Q.5 The resistance of a conductor is 10 \(\Omega\) at 50⁰ C and 15 \(\Omega\) at 100 ⁰ C. Calculate its resistance at 0 ⁰ C. ( Click here for the solution! ).

I-V graph of ohmic and non-ohmic conductors

Ohm's law is not correct in all situations. The conducting materials that obeys Ohm's law are called ohmic conductors. Examples are, copper, silver, etc. The conducting materials that doesnot obey Ohm's law are called non-ohmic conductors. Examples are, electrolytes, p-n junction diode, etc. Fig(4) represents the current-voltage graph of an ohmic conductor and Fig(5) represents the current-voltage graph of p-n junction diode.
Let's take a break for a thought here! Even in a non-ohmic conductor (p-n junction diode), if we measure the potential difference V across it, and current I through it, we can find that reistance, \(R=\frac{V}{I}\). However, the plot of V and I is not linear (which is the only essence of Ohm's law).

Fig(4): I-V graph of Ohmic resistor/conductor (a typical metal wire).

Fig(5): I-V graph of non-ohmic resistor/conductor (a p-n junction diode).

Determination of Resistance in series by applying the conservation of charge and energy:

Fig(6): Series combination of resistors.
Consider two resistors of resistances R₁ and R₂ connected in series. When this combination is connected to the two terminals of the battery (as in Fig(6)), same current (say I ) flows through the circuit which would otherwise violate the conservation of charge. According to the law of conservation of charge, all the charge that passes any point of the circuit in a given time interval must pass any other point in the same time interval (Charge is not lost, gained and accumulated anywhere).
Let V be the total potential difference across the combination. V₁ and V₂ be the potential difference (or potential drop) across the resistors of resistance R₁ and R₂ respectively.
Law of conservation of energy demands that the total work done on the charge q in the complete circuit is equal to zero (since no gain or lose of electrical potential energy is encountered in the complete circuit). So, the total potential differenceis equal to the potential difference across each resistors. \[V_1+V_2=V \text{... (i)}\] Now, using Ohm's law, \[\begin{align*} V_1&=IR_1\\ V_2&=IR_2\\ \end{align*}\] Equation (i) then becomes, \[\begin{align*} IR_1+IR_2&=IR_{eq}\\ \therefore R_{eq}&=R_1+R_2\\ \end{align*}\] R_eq is called the equivalent reistance of the combination which can be interpreted as the single resistor that replace the combination of resistors in the circuit.
Generealizing for any number of resistors in series, we can write \[R_{eq}=R_1 + R_2 + R_3 +...\] Thus, the equivalent resistance of any number of resistors in series equals the sum of their individual resistances.

Determination of Resistance in parallel by applying the conservation of charge and energy:

Fig(7): Parallel combination of resistors.
Consider two resistors of resistances R₁ and R₂ connected in parallel. When this combination is connected to the two terminals of the battery (as in Fig(7)), the current ( I ) splits at the junction into two parts I₁ and I₂. Law of conservation of charge demands that all charge entering the junction at a particular time must leave the junction. So, \[I=I_1+I_2 {\text{... (i)}}\] Since, the two resistors are connected across the same ends, the potential difference across the two ends must be equal to the potential difference across the battery (Law of conservation of energy). Using Ohm's law, we can write, \[\begin{align*} I_1&=\frac{V}{R_1}\\ I_2&=\frac{V}{R_2}\\ \end{align*}\] Equation (i) then becomes, \[\begin{align*} \frac{V}{R_1}+\frac{V}{R_2}&=\frac{V}{R_{eq}}\\ \frac{1}{R_1}+\frac{1}{R_2}&=\frac{1}{R_{eq}}\\ \therefore \frac{1}{R_{eq}}&=\frac{1}{R_1}+\frac{1}{R_2}\\ \end{align*}\] R _eq is called the equivalent resistance of the parallel combination of the resistors.
Generalizing for any number of resistors in parallel, we can write \[\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...\] Thus, the reciprocal of the equivalent reistance of any number of resistors in parallel equals the sum of the reciprocals of the individual resistances.

Q.6 Using the law of conservation of charge and energy, find the equivalent resistance for the number of resistors with resistance R₁, R₂, R₃.... connected in series and in parallel. ( See above! )

Electromotive force, internal resistance and terminal potential difference

In the Fig(8) below, we can see that the water comes out of the openenings at the top of the water fountain. Water then stage at different heights and ultimately projects downwards (towards the direction of decreasing gravitational energy) and collects in the basin in the bottom. Is the water poured out of the top of the fountain on its own? No, its actually the water pump that lifts the water to the top (towards increasing gravitational potential energy).

Fig(8): Water fountain (Source: gettyimages)
In an electric circuit there must be a device somewhere in the loop (a complete circuit) that acts like the water pump in a water fountain. In this device a charge travels from lower to higher potential energy (electrical potential energy) even though the electrostatic force is trying to push it from higher to lower potential energy. An example of such a device is a battery.

Fig(9): emf source with terminals a and b .
Consider two terminals a and b of a battery as in Fig(9). a , marked '+', is maintained at higher potential than b, marked '-'. An electric field \(\vec{E}\) is developed in the material from a to b , as shown. A charge (+q) within the source (battery) experiences an electric force \(\vec{F_e}=q\vec{E}\). But the battery also provides an additional influence, which is non-electrostatic force \(\vec{F_b}\). \(\vec{F_b}\) operating inside device, pushes charge from b to a , in a direction against the electrostatic force \(\vec{F_e}\). Thus \(\vec{F_b}\) maintains the potential difference between the terminals. If \(\vec{F_b}\) were not present, charge would flow between the terminals until the potential difference was zero.
In a steady state, the charge accumulation on a and b is such that \(F_b=F_e\). No further accumulation takes place.
If a charge q is taken from b to a , the work done by the battery force \(F_b\) is, \[W=F_b d\] where d is the distance between a and b . The work done by the battery force per unit charge is, \[\mathcal{E}=\frac{W}{q}=\frac{F_b d}{q}\] This quantity, \(\mathcal{E}\), is called the electromotive force (emf) of the battery. emf is not actually a force but a work done per unit charge .
If nothing is connected externally between a and b , \[\begin{align*} F_b&=F_e=qE\\ F_b d&= F_e d=qEd=qV\\ \end{align*}\] where \(V=Ed\) is the potential difference between two terminals of the battery. Thus, \[\frac{F_b d}{q}=V=\mathcal{E} \Longrightarrow \mathcal{E}=V\] Thus, emf of a battery equals the potential difference between its terminals when the terminal are not connected externally.

Any practical emf source has a resistance due to which \(\mathcal{E} \neq V\) when current flows through the source (battery). This reistance is called the internal resistance of the battery, denoted by r . As the current moves through r , it experiences an associated potential drop of Ir.

Fig(10): Practical emf source with internal resistance \(r\).
Thus, when a current is flowing through source from the negative terminal b to the positive terminal a , the potential difference between the terminals is, \[V=\mathcal{E} - Ir\] This potential is called the terminal potential difference and is less than \(\mathcal{E}\) because of the term Ir . So for a real source of emf, \[V=\mathcal{E} \hspace{0.1cm} \text{when} \hspace{0.1cm} I=0\]

Circuit Formula

Circuit formula gives the relation between emf, terminal potential difference and internal resistance of a cell.

Fig(11): Closed loop for the determination of circuit formula.
Let us consider a cell of emf \(\mathcal{E}\) and internal resistance \(r\) is connected to a external resistance \( R\) in a circuit as in Fig(11). \( I \) be the current flowing through the circuit and \( V \) be the terminal potential difference.
We know, \[V=\mathcal{E} -Ir \hspace{0.1cm} \text{... (i)} \] Since the terminal potential difference is equal to the potential drop across the external resistor R , we can write , \[V=IR \hspace{0.1cm} \text{... (ii)}\] From (i) and (ii), \[\begin{align*} V&=\mathcal{E}-Ir\\ IR&=\mathcal{E}-Ir\\ \mathcal{E}&=I(R+r)\\ \therefore I&=\frac{\mathcal{E}}{R+r} \hspace{0.1cm} \text{... (iii)}\\ \end{align*}\] Also, from equation (i) and (ii), we see that \[I=\frac{\mathcal{E}-V}{r} \hspace{0.1cm} \text{... (iv)}\] From equation (iii) and (iv), we get, \[\begin{align*} \frac{E}{R+r}&=\frac{\mathcal{E}-V}{r}\\ \mathcal{E}r&=(\mathcal{E}-V)(R+r)\\ \mathcal{E}r&=\mathcal{E}R+\mathcal{E}r-VR-Vr\\ Vr&=(\mathcal{E}-V)R\\ \therefore r&=\left(\frac{\mathcal{E}-V}{V}\right) R \hspace{0.1cm} \text{... (v)}\\ \end{align*}\] Equation (v) gives the relation between emf, terminal potential difference and internal resistance of a cell.

Expression for the heat developed in a wire and power of an electrical circuit

Fig(12): Heat developed in a wire.
Consider a simple circuit with external resistor AB with resistance \(R\) is connected to a battery of emf \(\mathcal{E}\) through connecting wires as in Fig(12). If \(I\) be the current in the circuit, the terminal potential difference is \[V=IR \hspace{0.1cm} \text{... (i)}\] Let the charge \(q\) goes through the circuit in time \(t\). As this charge moves from A to B, the electric potential energy decreases by (since the charge is moving from higher potential to lower potential) \[\begin{align*} W&=qV\\ &=It.IR \hspace{0.1cm} \because I=\frac{q}{t}\\ &=I^2 R t \hspace{0.1cm} \text{... (ii)}\\ \end{align*}\] This loss in electric potential energy appears as increased thermal energy of the resistor. So, the heat developed in the wire is, \[H=I^2 Rt\] Now, the power developed is, \[P=\frac{W}{t}=\frac{I^2 Rt}{t}=I^2 R\] Usig Ohm's law (\(V=IR\)), we can write, \[P=IV \hspace{0.1cm}\]

Q.7 Deduce an expression for the heat developed in a wire by the passage of an electric circuit. ( See the solution above! )

Q. 8 Two resistors of resistance 1000 \(\Omega\) and 2000 \(\Omega\) are joined in series with a 100 V supply. A voltmeter of internal resistance 4000 \(\Omega\) is connected to measure the potential difference across 1000 \(\Omega\) resistor. Calculate the reading shown by the voltmeter. ( Click here for the solution!)

Q. 9 A battery of emf 1.5 V has a terminal potential difference of 1.25 V when a resistor of 25 \(\Omega\) is joined to it. Calculate the current flowing, the internal resistance and terminal potential difference when a resistance of 10 \(\Omega\) replaces 25 \(\Omega\) resistor. ( Click here for the solution! )

Q. 10 In the given figure, when switch S is open, the voltmeter V reads 3.08 V. When the switch S is closed, the voltmeter reading drops to 2.97V, and the ammeter A reads 1.65 A. Find the emf, the internal resistance of the battery and the resistor R. Assume that the two meters are ideal. ( Click here for the solution! )

Q. 11 What is the potential difference across 100 \(\Omega\) resistor in the circuit given below? ( Click here for the solution! )

Q. 12 In the figure, the current through the 3 \(\Omega\) resistor is 0.8 \(\Omega\) A. Find the potential difference across the 4 \(\Omega\) resistor. ( Click here for the solution! )

Q. 13 Consider the figure below. The current through 6 \(\Omega\) resistor is 4 A in the direction shown. What are the currents through the 25 \(\Omega\) and 20 \(\Omega\) resistors? ( Click here for the solution! )

Q. 14 Two lamps rated 25 W - 220 V and 100 W - 220 V are connected to 220 V supply. Calculate the powers consumed by the lamps. ( Click here for the solution! )

Q. 15 An electric heating element to dissipate 480 watts on 240 V mains is to be made from nichrome wire of 1 mm diameter. Calculate the length of the wire required if the resistivity of nichrome is 1.1 \(\times\) 10\(^{-6}\) \(\Omega m\). ( Click here for the solution! )

Q. 16 An electric heating element to dissipate 1.2 Kw on 240 V mains is to be made from nichrome ribbon 1 mm wide and 0.05 mm thick. Calculate the length of the ribbon required if the resistivity of nichrome is 1.1 \(\times\) 10\(^{-6}\) \(\Omega m\). ( Click here for the solution! )

Q. 17 Twelve cells each of emf 2 V and of internal resistance 0.5 \(\Omega\) are arranged in a battery of n rows and an external resistance 0.4 \(\Omega\) is connected to the poles of the battery. Estimate the current flowing through the resistance in terms of n. ( Click here for the solution! )

Short answer question solution

Q. 1 A proton beam is going from east to west. Is there an electric current? If yes, in what direction?
Solution:
Electric current is the rate of flow of charge. Conventionally, the direction of current is in the direction of motion of positive charge. Since, proton is a positive charge, the current is thus in its direction i.e., from east to west.

Q. 2 In an electrolyte, the positive ions move from left to right and the negative ions from right to left. Is there a net current? If yes, in what direction?
Solution:
Here, the direction of negative charge is towards right to left and that of positive charge is from left to right. Since the direction of current is same as the direction of motion of positive charge, the current due to the positive ions is to the right and as the current due to the negative ions is opposite to the direction of motion of negative ions and is also thus directed from left to right. Hence, the net current is directed from left to right.

Q. 3 Two copper wires of different diameter are joined end to end. What will be the drift velocity of electrons in these two wires after passing current?
Solution:
Let, \(d_1\) and \(d_2\) be the diameters of two copper wires having respective drift velocity \(v_{d_1}\) and \(v_{d_2}\). Since they are joined end to end, same current (say \(I\) ) flows through each of them. We have, \[I=v_d e n A\] Thus, \[\begin{align*} v_{d_1} e n A_1&=v_{d_2} e n A_2\\ v_{d_1}A_1&=v_{d_2} A_2\\ \frac{v_{d_1}}{v_{d_2}}&=\frac{A_2}{A_1}\\ \frac{v_{d_1}}{v_{d_2}}&=\frac{\pi \frac{{d_2}^2}{4}} {\pi \frac{{d_1}^2}{4}}\\ \frac{v_{d_1}}{v_{d_2}}&=\frac{{d_2}^2}{{d_1}^2}\\ \end{align*}\] From the above expression, it is seen that \[v_d \propto \frac{1}{d^2}\] Thus, the drift velocity of electrons in wire having larger diameter will be smaller and that in wire having smaller diameter will be greater.

Q. 4 A wire is stretched to double its length. What happens to its resistivity and resistance?
Solution:
Resistivity (\(\rho\)) of a conductor depends only on the material of the conductor. Hence, it does not change on doubling the length of wire.
Let, \(l\) be the initial length and \(l'=2l\) be the double the length of the wire. Initially, the resistance is, \[R=\frac{\rho l}{A}\] It can also be written in terms of volume \(V=Al\) as, \[R=\frac{\rho l^2}{V} \hspace{0.1cm} \text{... (i)}\] At length \(l'\), the resistance becomes, \[R'=\frac{\rho l'^2}{V} \hspace{0.1cm} \text{.... (ii)}\] Here, the volume does not change on doubling the length of wire because as we double the length, the cross section area of wire decreases making the volume constant.
Dividing equation (ii) by (i), we get, \[\begin{align*} \frac{R'}{R}&=\frac{l'^2}{l^2}\\ \frac{R'}{R}&=\frac{(2l)^2}{l^2}\\ \frac{R'}{R}&=4\\ \therefore R'&=4 \times R\\ \end{align*}\] Hence, the resistance of wire increases by 4 times on doubling the length of wire.

Q. 5 Batteries are always labeled with their emf; for instance, an AAflashlight battery is labeled “1.5 volts.” Would it also be appropriate to put a label on batteries stating how much current they provide? Why or why not?
Solution:
For any real battery of emf with internal resistance \(r\), the terminal potential difference is calculated as, \[V=\mathcal{E}-Ir\] Here, \(V=IR\) and thus, \[\begin{align*} IR&=\mathcal{E}-Ir\\ I(R+r)&=\mathcal{E}\\ I&=\frac{\mathcal{E}}{R+r}\\ \end{align*}\] From this relation, we can see that the battery will produce less current when it is connected with larger external resistance. Similarly, when the battery is connected to smaller external resistance, it produces larger current. Thus, it is not appropriate to label the current they provide to the circuit.

Q. 6 The drift speed of electrons is very slow, often on the order of 10\(^{-4}\) m/s. Given that the electrons move so slowly, why the light comes on immediately when you turn on the switch of a flashlight?
Solution:
The electric field is set up in the wire with a speed approaching the speed of light, and electrons start to move all along the wire at very nearly the same time. The time that it takes any individual electron to get from the switch to the light bulb isn’t really relevant. Due to this reason, light comes immediately when we turn on the switch of a flash light.

Q. 7 Why does an electric light bulb nearly always burn out just as you turn on the light, almost never while the light is shining?
Solution:
An electric light bulb has an resistance (say \(R\)) and as we turn on the light, current passes through it and the heat (\(H=I^2 Rt\)) is developed in the bulb abruptly which expands the filament wire rapidly due to which it might burn instantly. But, if it is shining, the temperature of filament is relatively stable with no sudden expansion or contraction as in the initial case. Due to this reason, an electric light bulb nearly burns out as we turn on the light and almost never while the light is shining.

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