Figure shows a part of an electric circuit. The potentials at points a, b, and c are 30 V, 12 V, and 2 V respectively. The current through the 20 resistor will be:

Given: Potentials at a = 30 V, b = 12 V, and c = 2 V. The resistors are 10, 20, and 30. Find: The current through the 20 resistor. Let potential of the junction be x volts. Using junction law, i1 + i2 + i3 = 0 From the solution, x - 30 = x - 1210 and x - 2 = x - 1220 Solving for x, x = 20 V Therefore, current through the 20 resistor is 20 - 1220 = 0.4 A Therefore, the current through the 20 resistor is 0.4 A. The correct option is C.

Figure shows a part of an electric circuit. The potentials at poi… — Solution | JEE Physics

MCQMediumJEE 2023Kirchhoff's Laws & Circuits

Figure shows a part of an electric circuit. The potentials at points $a$ , $b$ , and $c$ are $30 \, \text{V}$ , $12 \, \text{V}$ , and $2 \, \text{V}$ respectively. The current through the $20 \, \Omega$ resistor will be:

Circuit with point a connected through a 10 ohm resistor to a junction, which splits into an upper 20 ohm branch ending at b and a lower 30 ohm branch ending at c, with current directions shown left to right.

A
$1.0 \, \text{A}$
B
$0.2 \, \text{A}$
C
$0.4 \, \text{A}$
D
$0.6 \, \text{A}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Potentials at $a = 30 \, \text{V}$ , $b = 12 \, \text{V}$ , and $c = 2 \, \text{V}$ . The resistors are $10 \, \Omega$ , $20 \, \Omega$ , and $30 \, \Omega$ .

Find: The current through the $20 \, \Omega$ resistor.

Let potential of the junction be $x$ volts. Using junction law,

i_1 + i_2 + i_3 = 0

From the solution,

x - 30 = \frac{x - 12}{10}

and

x - 2 = \frac{x - 12}{20}

Solving for $x$ ,

x = 20 \, \text{V}

Therefore, current through the $20 \, \Omega$ resistor is

\frac{20 - 12}{20} = 0.4 \, \text{A}

Therefore, the current through the $20 \, \Omega$ resistor is $0.4 \, \text{A}$ . The correct option is C.

Common mistakes

Using the potential difference across the wrong resistor branch is a common mistake. The $20 \, \Omega$ resistor is connected between the junction and point $b$ , so its current must be found using the junction potential and $12 \, \text{V}$ , not the potentials at $a$ and $c$ .
Assuming the three resistors are in simple series is incorrect because the circuit splits at the junction into two branches. After the $10 \, \Omega$ resistor, current divides between the $20 \, \Omega$ and $30 \, \Omega$ paths.
Taking the given point potentials as direct voltage drops across each resistor is wrong. Potentials are measured with respect to a common reference, so the voltage across a resistor is the difference between the potentials at its two ends.

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Answer

Step-by-step solution

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