What is the net resistance between points A and F in the circuit shown in fig.(a)?

Fig. (a)

Fig. (b)

Fig. (c)

Fig. (d)

Solution

Firstly, considering the parallel combination BC.

The resistance of this combination is given by

\frac{1}{R} = \frac{1}{30} + \frac{1}{15} = \frac{3}{30} R = 10 Ω

Substitute 10 Ω for the combination and redraw the circuit as shown in fig.(b).

Now, the series combination BD can be replaced by a single resistor of resistance
R = 10 + 2 = 12 Ω

Now the circuit can be redrawn as in fig. (c).

Combination AE in the parallel arrangement can be replaced by a single resistor of resistance given by

\frac{1}{R} = \frac{1}{12} + \frac{1}{60} = \frac{6}{60} R = 10 Ω

The circuit can again redrawn as shown in fig.(d).

The resistance between points A and F (series) is given by
R = 10 + 7 = 17 Ω.

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Current Electricity

What is the net resistance between points A and F in the circuit shown in fig.(a)?

Fig. (a)

Fig. (b)

Fig. (c)

Fig. (d)

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Current Electricity

What is the net resistance between points A and F in the circuit shown in fig.(a)?Fig. (a)Fig. (b)Fig. (c)Fig. (d)

Some More Questions From Current Electricity Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

What is the net resistance between points A and F in the circuit shown in fig.(a)?

Fig. (a)

Fig. (b)

Fig. (c)

Fig. (d)