Skip to main content

In the figure, the current through the 3 Ω resistor is 0.8 Ω A. Find the potential difference across the 4 Ω resistor.

-

In the figure, the current through the 3 Ω resistor is 0.8 Ω A. Find the potential difference across the 4 Ω resistor.
Solution:
Let, \(R_1\) = 3 \(\Omega\)
\(R_2\) = 6 \(\Omega\)
\(R_3\) = 4 \(\Omega\)
\(I_1\)=0.8 A

Since R1 and R2 are parallel to each other, the potential difference across each of these are equal.
p.d across R1 = p.d. across R2
Then, \[\begin{align*} I_1\times R_1&=I_2\times R_2\\ 0.8 \times 3&=I_2 \times 6\\ \therefore I_2&=0.4 \hspace{0.1cm} A \end{align*}\] The total current is then, \[I=I_1+I_2=1.2 \hspace{0.1cm} A\] So, the potential difference across the 4 Ω resistor is, \[\begin{align*} &=I\times 4\\ &=1.2 \times 4\\ &=4.8 \hspace{0.1cm} V\\ \end{align*}\]

Comments

Post a Comment

Popular Posts

Class 11 NEB model question solution 2077 | Physics | Complete explanation and notes

-
Image
Model Question Solution Group 'A' 1. A metre rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimeter. How should this result be recorded in a table of results? (a) 0.2000 m (b) 0.200 m (c) 0.20 m (d) 0.2 m Explanation: From the question, it is obvious that the least count of the metre scale is 1 mm as it measures upto the millimeter. Since, 1 mm = 0.001 m . So, result should have three significant figures. Thus, \[\begin{align*} 20 cm &= \frac{20}{100}\\ \therefore 20 cm &=0.200 \hspace{0.01cm} m\\ \end{align*}\] 2. Forces are applied to a rigid body. The forces all act in the same plane. In which diagram is the body in equilibrium? (b) Explanation: In this case, the downward force has balanced the upward force and hence the body is in equilibrium. 3. An athlete makes a long jump a
Read more

Electromagnetic induction | Class 12 NEB Physics | Numerical problem solution

-
The magnetic flux passing perpendicular to the plane of coil is given by φ = 4t 2 +5t+2 where φ is in weber and t is in second. Calculate the magnitude of instantaneous emf induced in the coil when t = 2 sec. Solution: Given, magnetic flux, φ = 4t 2 +5t+2 time, t = 2 sec induced emf, ε = ? From Faraday's law, \[\begin{align*} \epsilon& = -\frac{d\phi}{dt}\\ &=-\frac{d(4t^2+5t+2)}{dt}\\ &=-(8t+5)\\ \end{align*}\] At t = 2 sec, the induced emf is , \[\begin{align*} \epsilon&=-(8 \times 2 + 5)\\ \therefore \epsilon& = -21\hspace{0.1cm}\text{V}\\ \end{align*}\] Thus, the magnitude of induced emf is 21 V. A straight conductor of length 25 cm is moving perpendicular to its length with a uniform speed of 10 m/s making an angle of 45 0 with a uniform magnetic field of 10 T. Calcuate the emf induced across its length. Solution: Given, length, l = 25 cm = 0.25 m sp
Read more

Electrons | Complete notes with short answer questions and numerical problem solutions | Class 12 NEB Physics

-
Image
Electrons Introduction and background Electron is the lightest subatomic particle. It carries a charge of -1.602176634 × 10 -19 C, which is considered the fundamental or basic unit of electric charge. The mass of an electron is 9.1093837015 × 10 -19 kg which is \(\frac{1}{1836}\) times the mass of a proton (a positively charged constituent of an atom). Discovery of electrons: The discharge tube experiment played an important role in the discovery of electrons. While studying the properties of cathode rays (1897 A.D.), J.J. Thomson ( Nobel prize for Physics, 1906 ) found that the cathode rays are deflected in electric and magnetic fields which led scientists to believe that the cathode rays are made of tiny negatively charged particles. ( Note! In this chapter, we will learn how the electrons get affected by the electric and magnetic fields so you would know why scientists came in conclusion to identify the discovered particle as negatively charged el
Read more