Question

In the circuit shown in figure, circuit is closed at time $t=0$. At time $t=\ln \left(2\right)$ second

• A. rate of energy supplied by the battery is $16J/s$
• B. rate of heat dissipated across resistance is $8J/s$
• C. rate of heat dissipated across resistance is $16J/s$
• D. $V_a-V_b=4V$

Answer 1

Answer: (A), (B), (D)

The correct options are
A rate of energy supplied by the battery is $16J/s$
B rate of heat dissipated across resistance is $8J/s$
D $V_a-V_b=4V$

${\tau }_L=\frac{L}{R}=\frac{2}{2}=1s$

$t_{\frac{1}{2}}=\left(\ln 2\right){\tau }_L=\left(\ln 2\right)s$

Hence the given time is half life time.

$\therefore$ $i=\frac{i_0}{2}=\frac{8/2}{2}=2A$

Rate of energy supplied by battery

$=Ei=8\times 2=16J/s$

$P_R=i^{2}R={\left(2\right)}^2\left(2\right)=8J/s$

$V_a-V_b=E-iR=8-2\times 2=4V$