1. In the figure given below, the value of the source voltage is ___________
a) 12 V
b) 24 V
c) 30 V
d) 44 V
Explanation: By applying KCL, \(\frac{V_P-E}{6} + \frac{V_P}{6}\) – 1 = 0
Or, 2 VP – E = 6
Where, (\(\frac{-V_P+E}{6}\)) = 2
E – VP = 12
VP = 18 V
E = 30V.
2. In the figure given below, the value of Resistance R by Thevenin Theorem is ___________
a) 10
b) 20
c) 30
d) 40
Explanation: \(\frac{V_P-100}{10} + \frac{V_P}{10}\) + 2 = 0
Or, 2VP – 100 + 20 = 0
VP = 80/2 = 40V
R = 20Ω.
3. In the figure given below, the Thevenin’s equivalent pair, as seen at the terminals P-Q, is given by __________
a) 2 V and 5 Ω
b) 2 V and 7.5 Ω
c) 4 V and 5 Ω
d) 4 V and 7.5 Ω
Explanation: For finding VTH,
VTH = \(\frac{4 ×10}{10+10}\) = 2V
For finding RTH,
RTH = 10 || 10
= \(\frac{10×10}{10+10}\) = 5 Ω.
4. The Thevenin equivalent impedance Z between the nodes P and Q in the following circuit is __________
a) 1
b) 1 + s + \(\frac{1}{s}\)
c) 2 + s + \(\frac{1}{s}\)
d) 3 + s + \(\frac{1}{s}\)
Explanation: To calculate the Thevenin resistance, all the current sources get open-circuited and voltage source short-circuited.
∴ RTH = (\(\frac{1}{s}\) + 1) || (1+s)
= \(\frac{\left(\frac{1}{s} + 1\right)×(1+s)}{\left(\frac{1}{s} + 1\right)+(1+s)}\)
= \(\frac{\frac{1}{s}+1+1+s}{\frac{1}{s}+1+1+s}\) = 1
So, RTH = 1.
5. While computing the Thevenin equivalent resistance and the Thevenin equivalent voltage, which of the following steps are undertaken?
a) Both the dependent and independent voltage sources are short-circuited and both the dependent and independent current sources are open-circuited
b) Both the dependent and independent voltage sources are open-circuited and both the dependent and independent current sources are short-circuited
c) The dependent voltage source is short-circuited keeping the independent voltage source untouched and the dependent current source is open-circuited keeping the independent current source untouched
d) The dependent voltage source is open-circuited keeping the independent voltage source untouched and the dependent current source is short-circuited keeping the independent current source untouched
Explanation: While computing the Thevenin equivalent voltage consisting of both dependent and independent sources, we first find the equivalent voltage called the Thevenin voltage by opening the two terminals. Then while computing the Thevenin equivalent resistance, we short-circuit the dependent voltage sources keeping the independent voltage sources untouched and open-circuiting the dependent current sources keeping the independent current sources untouched.
6. The temperature coefficient of a metal as the temperature increases will ____________
a) Decreases
b) Increases
c) Remains unchanged
d) Increases and remains same
Explanation: We know that the temperature coefficient is,
Given by, α = \(\frac{α_0}{1 + α_0 t}\)
Since temperature is present at the denominator, so with increase in temperature t, the denominator increases and hence the fraction decreases.
So, temperature coefficient decreases.
7. Given a wire of resistance R Ω. The resistance of a wire of the same material and same weight and double the diameter is ___________
a) 0.5 R
b) 0.25 R
c) 0.125 R
d) 0.0625 R
Explanation: Since diameter is double, area of cross-section is four times and length is one-fourth.
It can be verified by the following equation,
R2 = \(\frac{\frac{ρl}{4}}{4A}\)
= \(\frac{ρl}{16 A} = \frac{R}{16}\).
8. The star equivalent resistance of 3 resistors having each resistance = 5 Ω is ____________
a) 1.5 Ω
b) 1.67 Ω
c) 3 Ω
d) 4.5 Ω
Explanation: We know that for star connection, REQ = \(\frac{R X R}{R+R+R}\)
Given R = 5 Ω
So, REQ = \(\frac{5 X 5}{5+5+5}\)
= \(\frac{25}{15}\) = 1.67 Ω.
9. The charge associated with a bulb rated as 20 W, 200 V and used for 10 minutes is ____________
a) 36 C
b) 60 C
c) 72 C
d) 50 C
Explanation: Charge Q= It
Given I = \(\frac{20}{200}\) = 0.1 A, t = 10 X 60 sec = 600 sec
So, Q = 0.1 X 600 = 60 C.
10. For a series RL circuit having L = 5 H, current = 1 A (at an instant). The energy stored in magnetic field is ___________
a) 3.6 J
b) 2.5 J
c) 1.5 J
d) 3 J
Explanation: We know that, Energy, E = 0.5 LI2
Or, E = 0.5 X 5 X 12 = 2.5 J.