First slide

Writing potential drop across a open line

Question

Resistances of 6 $Ω$ each are connected in the manner shown in fig. With the current 0.5 ampere as
shown in the figure, the potential difference Vp -V_Qis

Moderate

A

3.6 V
B

6.0 V
C

3.0 V
D

7.2 V

Solution

Equivalent resistance R of the three resistances connected in parallel is given by
$\frac{1}{R^{'}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} or R^{'} = 2 Ω$
The equivalent resistance R' of the second circuit
$\frac{1}{R^{''}} = \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{1}{4} or R^{''} = 4 Ω$
The resistances .R' and R' are connected in series. So, the equivalent resistance of the circuit i'e,, between A and B
R= 2+ 4= 6 $Ω$
Potential difference between. A and B
$= R \times i = 6 \times 0 \cdot 5 = 3 V$

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