The potential difference between point A and B is

$\frac{20}{7} V$

$\frac{40}{7} V$

$\frac{10}{7} V$

zero

Solution

zero

From the given figure, current through lower branch of resistances which are joined in series

$i_{1} = \frac{10}{4 + 3} = \frac{10}{7} amp$

Again current through upper branch of resistances which are also connected in series

$i_{2} = \frac{10}{8 + 6} = \frac{10}{14} amp Now according to the Kirchhoff' s voltage law, V_{B} - V_{A} = 8 \times i_{2} - 4 \times i_{1} = 8 \times \frac{10}{14} - 4 \times \frac{10}{7} V_{A} - V_{B} = 0$

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Current Electricity

The potential difference between point A and B is

$\frac{20}{7} V$

$\frac{40}{7} V$

$\frac{10}{7} V$

zero

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Current Electricity

The potential difference between point A and B is 207 V 407V 107V zero

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The potential difference between point A and B is

$\frac{20}{7} V$

$\frac{40}{7} V$

$\frac{10}{7} V$

zero