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Standard XII

Physics

Mixed Combination of Cells

A current of ...

Question

A current of 2 A flows in an electric cicuit as shown in figure. The potential difference $(V_{R} - V_{S})$ in volts ( $V_{R}$ and $V_{S}$ are potentials at R and S respectively) is

-4

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Solution

The correct option is C + 4
$\Rightarrow I t h r o u g h R = 1 A$
$t h r o u g h S = 1 A$
$V_{R} - V_{S} = (V_{P} - V_{S}) - (V_{P} - V_{R})$
$= (7 \times 1) - (3 \times 1)$
$= 4 V$

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A current of 2 A flows in an electric cicuit as shown in figure. The potential difference (VR−VS)in volts (VR and VS are potentials at R and S respectively) is

The correct option is C + 4⇒I through R=1Athrough S=1AVR−VS=(VP−VS)−(VP−VR) =(7×1)−(3×1) =4V

A current of 2 A flows in an electric cicuit as shown in figure. The potential difference $(V_{R} - V_{S})$ in volts ( $V_{R}$ and $V_{S}$ are potentials at R and S respectively) is

The correct option is C + 4
$\Rightarrow I t h r o u g h R = 1 A$
$t h r o u g h S = 1 A$
$V_{R} - V_{S} = (V_{P} - V_{S}) - (V_{P} - V_{R})$
$= (7 \times 1) - (3 \times 1)$
$= 4 V$