Question

In the circuit shown, the switch S is open for a long time and is closed at t = 0. The current i(t) for $t\ge 0^{+}$ is

• A. $i\left(t\right)=0.5-0.125e^{-1000t}A$
• B. $i\left(t\right)=1.5-0.125e^{-1000t}A$
• C. $i\left(t\right)=0.5-0.5e^{-1000t}A$
• D. $i\left(t\right)=0.375e^{-1000t}A$

Answer 1

The correct option is A $i\left(t\right)=0.5-0.125e^{-1000t}A$

Method 1:
At $t=0^{-}$, the circuit is shown below:

Current will divide equally in both $10\Omega$ resistors as shown in diagram.
At $t\ge 0^{+}$, the circuit in s-domain is shown below:

$[L=15mH,i(0^{+})=i(0^{-})=0.75A]$
The above circuit can be rearranged as given below:

Applying nodal analysis,
$\frac{V\left(s\right)-\frac{1.5}{s}}{10}+\frac{V\left(s\right)}{10}+\frac{V\left(s\right)+15\times {10}^{-3}\times 0.75}{10+15\times {10}^{-3}s}=0$
$\frac{V\left(s\right)}{10}+\frac{V\left(s\right)}{10}+\frac{V\left(s\right)}{10+15\times {10}^{-3}s}$
$=\frac{15}{10s}-\frac{15\times {10}^{-3}\times 0.75}{10+15\times {10}^{-3}s}$
$V\left(s\right)\left[\frac{10+15\times {10}^{-3}s+5}{5(10+15\times {10}^{-3}s)}\right]$
$=15\left[\frac{10+15\times {10}^{-3}s-10s\times {10}^{-3}\times 0.75}{10s(10+15\times {10}^{-3}s)}\right]$
$(1+{10}^{-3}s)V\left(s\right)=\frac{(10+7.5\times {10}^{-3}s)}{2s}\left[\frac{s+1000}{1000}\right]$
$V\left(s\right)=\frac{10000+7.5s}{2000s}$
$V\left(s\right)=\frac{10000+7.5s}{2s(s+1000)}$
$=\frac{10000}{2\times 1000s}+\frac{\{10000+7.5(-1000\left)\right\}}{2(-1000)(s+1000)}$
$=\frac{5}{s}-\frac{1.25}{s+1000}$
Taking inverse Laplace transform,
$v\left(t\right)=5-1.25e^{-1000t}V$
$i\left(t\right)=\frac{V\left(t\right)}{10}$
$=0.5-0.125e^{-1000t}A$
Method 2:
$i\left(0^{+}\right)=\frac{0.75}{2}=0.375A$
$i\left(\infty \right)=0.5A$
$i\left(t\right)=i\left(\infty \right)-\left\{i\right(\infty )-i(0^{+}\left)\right\}{e}^{-Rt/L}$
where, R = equivalent resistance seen across L with current source opened

$R=10+\left(10||10\right)=15\Omega$
$i\left(t\right)=0.5-\{0.5-0.375\}{e}^{-15t/15\times {10}^{-3}}$
$=0.5-0.125e^{-1000t}A$