Physics

Current Electricity and Circuits

377 Questions

Current electricity and circuits questions cover resistors, EMF, internal resistance, and power calculations in series and parallel configurations. Solving these builds a strong understanding of electrical principles and circuit analysis. These physics problems are highly relevant for technical and science aptitude tests.

Resistor combinationsPower dissipationEMF and internal resistanceAC circuit analysisOperational amplifiers

Current Electricity and Circuits Questions

Multiple choice

An amplifier without feedback has a voltage gain of 50, input resistance of 1 K$\Omega$ and output resistance of 2.5 K$\Omega$. The input resistance of the current-shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2, is
  1. $\dfrac{1}{11}$ K$\Omega$
  2. $\dfrac{1}{5}$ K$\Omega$
  3. 5 K$\Omega$
  4. 11 K$\Omega$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$$ \text{We have} R_i = 1K\Omega, \beta = 0.2, A =50 \\

Thus, \qquad R_{if} = \dfrac{R_i}{{(1+A\beta)}} = \dfrac{1}{11} K \Omega $$-

Multiple choice

Consider a 300 $\Omega$, quarter - wave long (at 1 GHz) transmission line as shown in figure. It is connected to a 10 V, 50 $\Omega$ source at one end and is left open circuited at the other end. The magnitude of the voltage at the open circuit end of the line is

  1. 10 V

  2. 5 V

  3. 60 V

  4. 60/7 V
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ \dfrac{V_L}{V_{in}} = \dfrac{Z_O}{Z-{in}} \\ or \qquad V_L = \dfrac{Z_O}{Z_{in}} V_{in} = \dfrac{10 \times 300} {50} = 60V

$

Multiple choice

A mast antenna consisting of a 50 metre long vertical conductor operates over a perfectly conducting ground plane. It is base-fed at a frequency of 600 kHz. The radiation resistance of the antenna in Ohms is
  1. $\dfrac{2\pi^2}{5}$
  2. $\dfrac{\pi^2}{5}$
  3. $\dfrac{4\pi^2}{5}$
  4. $20\pi^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since antenna is installed at conducting ground,

$$ R_rad = 80 \pi^2 (\dfrac{dl}{\lambda})^2 = 80 \pi^2 \left( \dfrac{50}{0.5 \times 10^3} \right)^2 = \dfrac{4\pi^2}{5} \Omega$$

Multiple choice

With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed: (i) 1$\Omega$connected at port B draws a current of 3 A (ii) 2.5 $\Omega$ connected at port B draws a current of 2 A With 10 V dc connected at port A, the current drawn by 7 $\Omega$connected at port B is

  1. 3/7 A

  2. 5/7

  3. 1A

  4. 9/7 A
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 Using Y-admittance model to figure out current for port B: I4 = Y21.V1 + Y22.V2 (i) from the 1st condition, -3 = Y21.10 + Y22.3 (since V2 = - I4 *RL = -(-3)*1 = 3V) (ii) from the 2nd condition, -2 = Y21.10 + Y22.5 (since V2 = - I4 *RL = -(-2)*2.5 = 5V) Solving these 2 equations, we get : Y21 = -9/20 and Y22 = 1/2 Hence, I4 = (-9/20).V1 + (1/2).V2 Given, V1 = 10V and V2 = (-I4 * 7) Substituting, We get I4 = -1A Ans : Current drawn by the 7$\Omega$ load resistor = -I4 = 1A

Multiple choice

With 10 V dc connected at port A in the linear non-reciprocal two-port network shown below, the following were observed:

(i) 1$\Omega$connected at port B draws a current of 3 A. (ii) 2.5 $\Omega$ connected at port B draws a current of 2 A.

For the same network, with 6 V dc connected at port A, 1 $\Omega$ connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is

  1. 6 V

  2. 7 V

  3. 8 V

  4. 9 V

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 on applying Thevenin's equation, for (i) and (ii), we get Vth = 3Rth +3.....(1) Vth = 2Rth +5...(2) subtracting (1) and (2), we get, Rth = 2ohms and Vth = 9V when input was 10V if the input is 6V then Vth = 7/3 * (2+1) =7V Vth = Vb=f(Vs) which implies Vth = mVs + c Vth(10V) = m(10) +C...(3)  Vth(6) = m(6)+ C...(4) Vth(7) = m(7) + C....(5) and Vth(8V) = m(8) + C.....(6) by substituting values in eq. (3 & 4) we get, 9 = 10m +C...(7) and 7 = 6m + C...(8) subtracting(7) and (8), we get, m = 0.5 and c = 4 now, Vth(8V) = m(8)+C = 8 volts