Example 1 - The switch in Fig. 1 has been in position A for a long time. At 𝑡 = 0 the
switch moves to B. Determine 𝑣(𝑡) for 𝑡 > 0 and calculate its value at 𝑡 = 1 𝑠 and
𝑡 = 4 𝑠.
Fig. 1
Solution –
For 𝑡 < 0 , the switch is at position A. The capacitor acts like an open circuit to dc, but
𝑣 is the same as the voltage across the 5 kΩ resistor.
Hence, the voltage across the capacitor just before 𝑡 = 0 is obtained by voltage
division as
𝑣 (0− ) =
5
(24) = 15𝑉
5+3
Now, we know that the voltage across the capacitor cannot change instantaneously.
∴ 𝑣 (0+ ) = 𝑣 (0− ) = 15𝑉
For 𝑡 > 0 , the switch is in position B. The Thevenin resistance connected to the
capacitor is RTh = 4 kΩ and the time constant is
τ = R Th C = 4 x 103 x 0.5 x 10−3 = 2 s
Since the capacitor acts like an open circuit to dc at steady state, 𝑣 (∞) = 30𝑉
𝑡
𝑡
𝑡
−
−
−
Thus, 𝑣 (𝑡 ) = 𝑣 (∞) + [𝑣 (0) − 𝑣 (∞)]𝑒 𝑅𝐶 = 30 + (15 − 30)𝑒 2 = 30 − 15𝑒 2
At t = 1, 𝑣 (1) = 30 − 15𝑒
−
1
2
= 20.9 𝑉
4
At t = 4, 𝑣 (4) = 30 − 15𝑒 −2 = 27.97 𝑉
Example 2 - Find 𝑖 (𝑡 ) in the circuit of Fig. 2 for 𝑡 > 0. Assume that the switch has
been closed for a long time.
Solution –
When 𝑡 < 0, the resistor is short-circuited, and the inductor acts like a short circuit.
The current through the inductor at 𝑡 = 0− (i.e., just before 𝑡 = 0) is
𝑖 (0− ) =
10
= 5𝐴
2
Fig. 2
Since the inductor current cannot change instantaneously,𝑖 (0+ ) = 𝑖 (0− ) = 5𝐴
When 𝑡 > 0 the switch is open. The 2 Ω and 3 Ω resistors are in series, so that
𝑖 ( ∞) =
10
= 2𝐴
2+3
The Thevenin resistance across the inductor terminals is RTh = 2 + 3 = 5 Ω.
Time constant, τ = L / RTh = 1/15 s.
𝑅
Thus,
𝑖 (𝑡 ) = 𝑖(∞) + [𝑖(0) − 𝑖 (∞)]𝑒 −( 𝐿 ) 𝑡
∴ 𝑖(𝑡) = 2 + [5 − 2]𝑒 −15 𝑡 = 2 + 3𝑒 −15 𝑡
Example 3 – In Fig. 3, the switch has been in position 1 for a long time and abruptly
changes to position 2 at t = 0. If time t is in seconds, find the capacitor voltage VC (in
volts) for t > 0.
Fig. 3
Solution - Voltage across capacitor is given by,
𝑡
𝑣 (𝑡 ) = 𝑣 (∞) + [𝑣 (0) − 𝑣 (∞)]𝑒 −𝑅𝐶
At < 0 : In steady state, capacitor behaves as an open circuit. Switch is at position ‘1’.
2
𝑣𝑐 (0− ) =
x 10 = 4 𝑉
2+3
From property of capacitor, 𝑣𝑐 (0− ) = 𝑣𝑐 (0+ ) = 4 𝑉
At > 0 : RTh τ = R Th C

In steady state, capacitor behaves as open circuit.
𝑣𝑐 (∞) = 2 x 5 = 10 𝑉
𝑡
𝑡
∴ 𝑣 (𝑡 ) = 10 + [4 − 10]𝑒 −6 x 0.1 = 10 − 6𝑒 −0.6
Problems
1. Find 𝑣 (𝑡 ) for 𝑡 > 0 in the circuit of Fig. 4. Assume the switch has been open for a
long time and is closed at 𝑡 = 0. Calculate 𝑣 (𝑡 ) at 𝑡 = 0.5.
Fig. 4
Answer: (9.375 + 5.625𝑒 −2𝑡 )𝑉 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0, 7.63 𝑉
2. The switch in Fig. 5 has been closed for a long time. It opens at 𝑡 = 0. Find 𝑖(𝑡)
for 𝑡 > 0.
Fig. 5
Answer: (4 + 2𝑒 −10𝑡 ) A 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
3. The switch in Fig. 6 opens at 𝑡 = 0. Find 𝑣𝑜 for 𝑡 > 0.
Fig. 6
−𝑡
Answer: (4𝑒 12 ) V
4. Calculate the capacitor voltage for 𝑡 < 0 and 𝑡 > 0 for the circuit shown in Fig. 7.
Fig. 7
−𝑡
Answer: 4 V for t < 0, (12 − 8𝑒 6 ) V for t > 0
5. Determine the inductor for 𝑡 < 0 and 𝑡 > 0 for the circuit shown in Fig. 8.
Fig. 8
−𝑡
Answer: 5 A, (5𝑒 2 ) A
References:
Fundamentals of Electrical Circuits, Charles K. Alexander and Matthew N.O.
Sadiku, 5Th edition, McGraw Hill Education.