142. An 8 µF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the initial charging current.
A. 100 µn
B. 500 µA
C. 400 µA
D. 1000 µA
Answer: C
In a series RC circuit, the initial charging current is:
I = V/R = 200/(0.5 × 106s)
= 400 × 10-6A = 400 µA.
143. An 8 µ F capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the time taken for the potential difference across the capacitor to grow to 160V.
A. 6.93 s
B. 7.77 s
C. 2.33 s
D. 3.22 s
Answer: A
We know from the previous explanations that the initial current is 400mA, and the time constant is 4s.
Substituting the values of capacitor voltage, initial voltage, initial current and time constant in the equation:
V = V0(1-e-t/RC)
Substituting V = 160V,
V0 = 200V,
RC = 4s we get,
t = 6.93s.
144. An 8 µF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the voltage in the capacitor 4s after the power is supplied.
A. 123.4 V
B. 126.4 V
C. 124.5 V
D. 132.5 V
Answer: B
We can get the value of the potential difference across the capacitor in 4s, from the following equation:
Vc = V(1-e-t /RC).
Substituting the values in the given equation, we get Vc = 126.4V.
145. An 8 µF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the current in the capacitor 4s after the power is supplied.
A. 79 µA
B. 68 µA
C. 48 µA
D. 74 µA
Answer: D
In the given question, the time constant is equal to the time taken = 4s.
Hence the value of current will be 37% of its initial value
= I = 0.37 × 200 = 74 µA.
146. A circuit has a resistance of 2 ohms connected in series with a capacitance of 6F. Calculate the discharging time constant.
A. 3
B. 1
C. 12
D. 8
Answer: C
The discharging time constant in a circuit consisting of a capacitor and resistor in series is the product of the resistance and capacitance
= 2 × 6 = 12.
147. What is the energy in a capacitor if the voltage is 5V and the charge is 10 C?
A. 25 J
B. 35 J
C. 54 J
D. 55 J
Answer: A
We know that Q/V = C.
Hence the value of capacitance is 2F.
U = (Q × V)/2 = (10 × 5)/2 = 25 J.
151. A CR network is one which consists of _________
A. A capacitor and resistor connected in parallel
B. A capacitor and resistor connected in series
C. A network consisting of a capacitor only
D. A network consisting of a resistor only
Answer: B
A CR network is one that consists of a capacitor connected in series with a resistor. The capacitor discharges or charges through the resistor.
152. At DC capacitor acts as _________
A. Open circuit
B. Short circuit
C. Resistor
D. Inductor
Answer: A
Capacitive Reactance XC = 1/(2πfC).
For DC, f = 0 so, XC becomes infinite.
Hence for dc, the capacitor acts as an open circuit.
153. In an RC series circuit, when the switch is closed and the circuit is complete, what is the response?
A. Response does not vary with time
B. Decays with time
C. Increases with time
D. First increases, then decrease
Answer: B
In an RC series circuit, the response decays with time because according to the equation, there is an exponential decrease in the response.
154. If the switch is closed at t = 0, what is the current in the circuit?
A. 0A
B. 10A
C. 20A
D. Infinity
Answer: B
As soon as the switch is closed at t = 0, the capacitor acts as a short circuit. The current in the circuit is:
I = V/R = 100/10 = 10A.
155. Calculate the voltage across the capacitor at t = 0.
A. 0V
B. 10V
C. 20V
D. Infinity
Answer: A
When the switch is closed at t = 0, the capacitor has no voltage across it since it has not been charged. The capacitor acts as a short circuit and the voltage across it is zero.
156. Calculate di(0)/dt if the switch is closed at t = 0.
A. -9.9A/s
B. -10A/s
C. 0A/s
D. -0.1A/s
Answer: D
Applying KVL to the given circuit, we get:
i = i0e-t/RC = (100/10)e-t/100
i = 10 e-t/100
di/dt = -(10/100) e-t/100
di(0)/dt = -0.1A/s.
157. Calculate d2i(0)/dt2 from the given circuit.
A. 10-6A/s2
B. 10-3A/s2
C. 106A/s2
D. 103A/s2
Answer: B
Applying KVL to the given circuit, we get:
100+10i(0)+1/10 × ∫(i(0)dt) = 0
Differentiating once, we get:
10di(0)/dt+1/10 × i.
Differentiating once again, we get:
10d2i(0)/dt2+10di(0)/dt = 0.
Substituting the values of di/dt from the previous explanation, we get d2i(0)/dt2 = 10-3A/s2.
158. The current equation for the given circuit is?
A. i = 10e(-0.01)t A
B. i = 10e(0.01)t A
C. i = 10e(-0.001)t A
D. i = 100e(-0.01)t A
Answer: A
The KVL equation is:
100+10i(0)+1/10 × ∫(i(0)dt) = 0
On applying a Laplace transform to this equation, we get:
100/s = I(s)/10s+10I(s)
Solving the equation, we get:
i = 10e(-0.01)t A.
159. The expression for the current in an RC circuit is?
A. i = (V/R)et/RC
B. i = (V/R)e-t/RC
C. i = (V/R)(1-e-t/RC)
D. i = (V/R) (1-et/RC)
Answer: B
Applying KVL to the given circuit, we get:
i = i0e-t/RC = (100/10)e-t/100
i = 10 e-t/100.
160. What is the voltage in the resistor as soon as the switch is closed at t = 0?
A. 0V
B. Infinity
C. 220V
D. Insufficient information provided
Answer: C
As soon as the switch is closed at t = 0, there is no charge in the capacitor, hence the voltage across the capacitor is zero and all the 220V voltage is the voltage across the resistor.
