Two batteries of emf 3 V and 6 V with internal resistance 2 - and 4 - are connected in a circuit with resistance of 10- as shown in figure- The current and potential difference between the points P and Q are

The correct option is D 3 16 A and 15 8 V Applying Kirchhoff's voltage law in the given loop, 4l+6 3 2l 10l=0 16l= 3 l= 3 16 A ...

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Byju's Answer

Standard XII

Physics

Mixed Combination of Cells

Two batteries...

Question

Two batteries of emf $3 V$ and $6 V$ with internal resistance $2 Ω$ and $4 Ω$ are connected in a circuit with resistance of $10 Ω$ as shown in figure. The current and potential difference between the points $P$ and $Q$ are

\frac{3}{16} A

and

\frac{8}{15} V

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\frac{16}{3} A

and

\frac{15}{8} V

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\frac{3}{16} A

and

8 V

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\frac{3}{16} A

and

\frac{15}{8} V

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Solution

The correct option is D $\frac{3}{16} A$ and $\frac{15}{8} V$
Applying Kirchhoff's voltage law in the given loop,
$- 4 l + 6 - 3 - 2 l - 10 l = 0$
$- 16 l = - 3$
$l = \frac{3}{16} A$
$∵$ Potential difference across $P Q = \frac{3}{16} \times 10 = \frac{15}{8} V$

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Two batteries of emf 3V and 6V with internal resistance 2Ω and 4Ω are connected in a circuit with resistance of 10Ω as shown in figure. The current and potential difference between the points P and Q are

The correct option is D 316A and 158VApplying Kirchhoff's voltage law in the given loop,−4l+6−3−2l−10l=0−16l=−3l=316A∵ Potential difference across PQ=316×10=158V

Two batteries of emf $3 V$ and $6 V$ with internal resistance $2 Ω$ and $4 Ω$ are connected in a circuit with resistance of $10 Ω$ as shown in figure. The current and potential difference between the points $P$ and $Q$ are

The correct option is D $\frac{3}{16} A$ and $\frac{15}{8} V$
Applying Kirchhoff's voltage law in the given loop,
$- 4 l + 6 - 3 - 2 l - 10 l = 0$
$- 16 l = - 3$
$l = \frac{3}{16} A$
$∵$ Potential difference across $P Q = \frac{3}{16} \times 10 = \frac{15}{8} V$