A potential divider circuit is shown in the figure. The output vo… — Solution | JEE Physics

MCQEasyJEE 2024Kirchhoff's Laws & Circuits

A potential divider circuit is shown in the figure. The output voltage ( $V_0$ ) is:

A
$4 \, \text{V}$
B
$2 \, \text{mV}$
C
$0.5 \, \text{V}$
D
$12 \, \text{mV}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Input voltage is $V_{in} = 4 \, \text{V}$ . The series circuit has one resistor of $3.3 \, \text{k}\Omega$ and seven resistors of $100 \, \Omega$ each. The output voltage $V_0$ is measured across the last five $100 \, \Omega$ resistors.

Find: The value of $V_0$ .

Identify principle: Use the voltage divider rule for a series circuit.

V_{out} = V_{in} \times \frac{R_{out}}{R_{total}}

Convert $3.3 \, \text{k}\Omega$ into ohms:

3.3 \, \text{k}\Omega = 3300 \, \Omega

Total resistance of the circuit:

R_{total} = 3300 \, \Omega + (7 \times 100 \, \Omega)

R_{total} = 3300 \, \Omega + 700 \, \Omega = 4000 \, \Omega

Resistance across which output is measured:

R_{out} = 5 \times 100 \, \Omega = 500 \, \Omega

Apply the voltage divider formula:

V_0 = 4 \, \text{V} \times \frac{500 \, \Omega}{4000 \, \Omega}

V_0 = 4 \times \frac{500}{4000}

V_0 = 4 \times \frac{5}{40} = 4 \times \frac{1}{8}

V_0 = 0.5 \, \text{V}

Therefore, the output voltage is $0.5 \, \text{V}$ and the correct option is C.

Current Method

Given: Total input voltage is $4 \, \text{V}$ . Total series resistance is $4000 \, \Omega$ , and the output is across $500 \, \Omega$ .

Find: The output voltage $V_0$ .

First calculate the circuit current:

i = \frac{4}{4000} = \frac{1}{1000} \, \text{A}

Now use $V = iR$ across the $500 \, \Omega$ portion:

V_0 = \frac{1}{1000} \times 500 = 0.5 \, \text{V}

Therefore, the output voltage is $0.5 \, \text{V}$ .

Common mistakes

Using the full series resistance as the output resistance. This is wrong because $V_0$ is measured only across the last five $100 \, \Omega$ resistors. Use $R_{out} = 500 \, \Omega$ , not $4000 \, \Omega$ .
Forgetting to convert $3.3 \, \text{k}\Omega$ into ohms. This is wrong because adding $3.3$ directly to resistances in ohms gives an incorrect total. Convert first: $3.3 \, \text{k}\Omega = 3300 \, \Omega$ .
Counting the number of $100 \, \Omega$ resistors incorrectly in the output section. This is wrong because the solution states that $V_0$ is across five such resistors. Count the resistors across the output terminals carefully before applying the divider rule.

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