Current Electricity Ques 63
- 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 $ $\Omega$
(b) $8 $ $\Omega$
(c) $1 $ $\Omega$
(d) $16 $ $\Omega$
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Answer:
Correct Answer: 63.(b)
Solution:
Formula:
- 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 $
$\therefore$ Equivalent circuit is
$\therefore$ 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} $ $\quad$ …….(i)
According to question, power consumed by the network, $P=4 W$.
From Eq. (i), we get
$ \therefore \quad \frac{16 \times 16}{8 R}=4 \Rightarrow R=\frac{16 \times 16}{8 \times 4}=8$ $ \Omega $
BETA
