SATHEE: Current Electricity 3 Question 2

Current Electricity 3 Question 2

2. The resistive network shown below is connected to a DC source of $16 V$ . The power consumed by the network is $4 W$ . The value of $R$ is

(2019 Main, 12 April I)

(a) $6 Ω$

(b) $8 Ω$

(d) $16 Ω$

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Solution:

Given circuit is

Equivalent resistance of part $A$ ,

$R_{A} = \frac{4 R \times 4 R}{4 R + 4 R} = 2 R$

Equivalent resistance of part $B$ ,

$R_{B} = \frac{6 R \times 12 R}{6 R + 12 R} = \frac{72}{18} R = 4 R$

$∴$ Equivalent circuit is

$∴$ Total resistance of the given network is

$R_{s} = 2 R + R + 4 R + R = 8 R$

As we know, power of the circuit,

$P = \frac{E^{2}}{R_{s}} = \frac{(16)^{2}}{8 R} = \frac{16 \times 16}{8 R}$

According to question, power consumed by the network, $P = 4 W$ .

From Eq. (i), we get

$∴ \frac{16 \times 16}{8 R} = 4 \Rightarrow R = \frac{16 \times 16}{8 \times 4} = 8 Ω$

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Current Electricity 3 Question 2

2. The resistive network shown below is connected to a DC source of 16V. The power consumed by the network is 4W. The value of R is

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2. The resistive network shown below is connected to a DC source of $16 V$ . The power consumed by the network is $4 W$ . The value of $R$ is