Question:
In the circuit shown, the current through 8 ohm is same before and after connecting E. The value of E is:
In the circuit shown, the current through 8 ohm is same before and after connecting E. The value of E is:
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Whenever the current through a resistor remains unchanged, its voltage drop also remains unchanged because \(V=IR\).
Updated On: Sep 25, 2026
- \( 12 \text{ V} \)
- \( 6 \text{ V} \)
- \( 4 \text{ V} \)
- \( 2 \text{ V} \)
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The Correct Option is C
Solution and Explanation
Concept:
If the current through a resistor remains unchanged after a source is connected across it, then the potential difference across that resistor must also remain unchanged according to Ohm's law.
Step 1: Find the current through the \(8\Omega\) resistor before connecting \(E\).
Initially, the circuit contains a \(12\text{ V}\) source and three resistors \(6\Omega\), \(8\Omega\), and \(10\Omega\) in series. \[ R_{\text{eq}}=6+8+10=24\Omega \] Hence the circuit current is \[ I=\frac{12}{24}=0.5\text{ A} \] Since all elements are in series, the current through the \(8\Omega\) resistor is also \[ I_{8}=0.5\text{ A}. \]
Step 2: Calculate the voltage across the \(8\Omega\) resistor.
Using Ohm's law, \[ V_{8}=I_{8}\times 8 \] \[ V_{8}=0.5\times 8=4\text{ V}. \]
Step 3: Use the given condition.
The current through the \(8\Omega\) resistor remains unchanged after connecting the battery \(E\). Therefore, the voltage across the \(8\Omega\) resistor must also remain unchanged. Hence the source connected across the same terminals must provide \[ E=4\text{ V}. \] Therefore, \[ \boxed{E=4\text{ V}} \]
Step 1: Find the current through the \(8\Omega\) resistor before connecting \(E\).
Initially, the circuit contains a \(12\text{ V}\) source and three resistors \(6\Omega\), \(8\Omega\), and \(10\Omega\) in series. \[ R_{\text{eq}}=6+8+10=24\Omega \] Hence the circuit current is \[ I=\frac{12}{24}=0.5\text{ A} \] Since all elements are in series, the current through the \(8\Omega\) resistor is also \[ I_{8}=0.5\text{ A}. \]
Step 2: Calculate the voltage across the \(8\Omega\) resistor.
Using Ohm's law, \[ V_{8}=I_{8}\times 8 \] \[ V_{8}=0.5\times 8=4\text{ V}. \]
Step 3: Use the given condition.
The current through the \(8\Omega\) resistor remains unchanged after connecting the battery \(E\). Therefore, the voltage across the \(8\Omega\) resistor must also remain unchanged. Hence the source connected across the same terminals must provide \[ E=4\text{ V}. \] Therefore, \[ \boxed{E=4\text{ V}} \]
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