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Standard XII

Physics

Kirchhoff's Junction Law

A battery of ...

Question

A battery of enf 12 V and internal resistance $2 Ω$ is connected to a $4 Ω$ resistor. What will be the current flowing through circuit? It a voltmeter is placed across the cell, what reading it will show?

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Solution

Current $I = \frac{12}{2 + 4} = 2 A$

The reading of the voltmeter will be equal to the potential difference between A and B.

$V_{A B} = V - 2 I = 12 - 2 \times 2$

$⟹ V_{A B} = 8 V$

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A battery of enf 12 V and internal resistance 2Ω is connected to a 4Ω resistor. What will be the current flowing through circuit? It a voltmeter is placed across the cell, what reading it will show?

Current I=122+4=2AThe reading of the voltmeter will be equal to the potential difference between A and B.VAB=V−2I=12−2×2⟹VAB=8V

A battery of enf 12 V and internal resistance $2 Ω$ is connected to a $4 Ω$ resistor. What will be the current flowing through circuit? It a voltmeter is placed across the cell, what reading it will show?

Current $I = \frac{12}{2 + 4} = 2 A$

The reading of the voltmeter will be equal to the potential difference between A and B.

$V_{A B} = V - 2 I = 12 - 2 \times 2$

$⟹ V_{A B} = 8 V$