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) $6 \Omega$
(b) $8 \Omega$
(c) $1 \Omega$
(d) $16 \Omega$
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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 $B$,
$$ R _B=\frac{6 R \times 12 R}{6 R+12 R}=\frac{72}{18} R=4 R $$
$\therefore$ 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,
$$ P=\frac{E^{2}}{R _s}=\frac{(16)^{2}}{8 R}=\frac{16 \times 16}{8 R} $$
According to question, power consumed by the network, $P=4 W$.
From Eq. (i), we get
$$ \therefore \quad \frac{16 \times 16}{8 R}=4 \Rightarrow R=\frac{16 \times 16}{8 \times 4}=8 \Omega $$
