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 physics current electricity meter bridge meter bridge and problems on it galvanometer

In a meter bridge an unknown resistance P is connected in the left gap and a $50 \Omega$ resistance in the right gap. Null point is obtained at x cm from the left end. The unknown resistance now shunted with an equal resistance. Find the value of the resistance in the right gap so that the null point is not shifted.

  1. $60 \Omega$
  2. $38 \Omega$
  3. $25 \Omega$
  4. $50 \Omega$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let P be the initial resistance. The null point condition is P/50 = x/(100-x). When P is shunted with an equal resistance P, the new resistance is P/2. To keep the null point at x, the right resistance R' must satisfy (P/2)/R' = x/(100-x). Comparing the two equations, P/50 = (P/2)/R', which gives R' = 25 ohms.

Multiple choice physics current electricity meter bridge meter bridge and problems on it galvanometer

Two resistors $R _1$ and $R _2$ are connected in the left gap and right gap of a meter bridge, and the null point is obtained at $20\;cm$ from the left. On interchanging the resistors in the two gaps. the null point shift by.

  1. $20\;cm$
  2. $40\;cm$
  3. $60\;cm$
  4. $80\;cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initially, R1/R2 = 20/80 = 1/4. After interchanging, R2/R1 = l/(100-l). Since R2/R1 = 4, l/(100-l) = 4, so l = 400 - 4l, 5l = 400, l = 80. The shift is 80 - 20 = 60 cm.

Multiple choice physics current electricity meter bridge meter bridge and problems on it galvanometer

Two equal resistances are connected in the gaps of a meter bridge. If the resistance in the left gap is increased by $10\%$, the balancing point shift :

  1. $10\%$ to right
  2. $10\%$ to left
  3. $9.6\%$ to right
  4. $4.8\%$ to right
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the Resistance be R,

Initially
$\dfrac{R}{R}  = \dfrac{l}{100-l}$
     $ \Rightarrow  l = 50 cm$
After 10% increase it is,

$ \Rightarrow  \dfrac{1.1R}{R} = \dfrac{l}{100-l}$

$ \Rightarrow  110  = 2.1 l$

$ \Rightarrow  l = \dfrac{110}{2.1} = 52.38$

$ \Rightarrow  \dfrac{\Delta l}{l} \times 100 = \dfrac{2.38}{50}\times 100  \approx 4.8$%
Multiple choice physics current electricity meter bridge meter bridge and problems on it galvanometer

In Wheatstone's bridge $  P=9  $ ohm, $  Q=11  $ ohm, $  R=4  $ ohm and $  S=6  $ ohm. How much resistance must be put in parallel to the resistance $  S  $ to balance the bridge

  1. $24 ohm$
  2. $ \frac{44}{9} ohm$
  3. $26.4 \mathrm{ohm} $
  4. $18.7 ohm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a balanced Wheatstone bridge, P/Q = R/S'. Here P/Q = 9/11. R/S' = 4/S'. So 9/11 = 4/S', S' = 44/9. The original S is 6. To get 44/9, we add a parallel resistance x: (6*x)/(6+x) = 44/9. 54x = 264 + 44x, 10x = 264, x = 26.4 ohms.

Multiple choice physics current electricity meter bridge meter bridge and problems on it galvanometer

In specific resistance measurement of a wire using a meter bridge, the key k in the main circuit is kept open when we are not taking readings. The reason is

  1. the emf of cell will decrease.

  2. the value of resistance will change due to joule heating effect.

  3. the galvanometer will stop working.

  4. none of these.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The key k in the main circuit is kept open when we are not taking the readings because the value of resistance will change due to joule heating effect. When an electric current passes through the resistor, heat is generated which rises the temperature of resistor which in result changes the resistance of resistor.

Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

A resistance $R\Omega$ is connected in series with capacitance $C$ Farad value of impedance of the circuit is $10\Omega$ and $R=6\Omega$ so, find the power factor of circuit.

  1. $0.4$
  2. $0.6$
  3. $0.67$
  4. $0.9$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In an RC circuit, the impedance Z = sqrt(R^2 + Xc^2). Given Z = 10 and R = 6, then 100 = 36 + Xc^2, so Xc = 8. The power factor is cos(phi) = R/Z = 6/10 = 0.6.

Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

In an AC circuit $V$ and $I$ are given by $V=100\sin{\left(100t\right)}$volt, $I=100\sin{\left(100t+\dfrac{\pi}{3}\right)}$amp the power dissipated in the circuit is

  1. $5.0\ kW$
  2. $2.5\ kW$
  3. $1.25\ kW$
  4. $zero$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$P={V} _{rms}\times{I} _{rms}\times\cos{\phi}$
$=\dfrac{{V} _{\circ}{I} _{\circ}}{2}\cos{\phi}$
$=\dfrac{100\times 100}{2}\times \dfrac{1}{2}$
$=2500$W
$=2.5$kW
Multiple choice physics capacitance capacitors in series combination of capacitors capacitors in parallel and series

The current in a contining a capacitance C and a resistance R in series over the applied voltage of frequency $\cfrac { \omega  }{ 2\pi  } $ by.

  1. ${ tan }^{ -1 }\left( \frac { 1 }{ \omega CR } \right) $
  2. ${ tan }^{ -1 }\left( \omega CR \right) $
  3. ${ tan }^{ -1 }\left( \omega \frac { 1 }{ R } \right) $
  4. ${ cos }^{ -1 }\left( \omega CR \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an RC series circuit, the phase angle phi between voltage and current is given by tan(phi) = Xc / R. Since Xc = 1 / (omega * C), the expression is tan(phi) = 1 / (omega * C * R). Thus, phi = tan^-1(1 / (omega * C * R)).

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

You measure the capacitor and inductor voltages in a driven RLC circuit, and find 10V for the rms capacitor voltage and 15V for the rms inductor voltage.

  1. $\omega = \omega _{res}$
  2. $\omega < \omega _{res}$
  3. $\omega > \omega _{res}$

  4. Can't be said
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
A/c to ques $V _{c}=10\ V$ and $V _{2}=15\ V$ 
Let frequency be $W$ capacitance be $C$ and inductance be $L$
$\Rightarrow iX _{C}=10$ and $iX _{L}=15$
$\Rightarrow \dfrac{i}{WL}=10$ and $i(WL)=15$
Dividing both
$\dfrac{\dfrac{i}{WC}}{i VWL}=\dfrac{10}{15}$
$\Rightarrow W^{2}=\dfrac{3/2}{LC}$
$W=\dfrac{\sqrt{1.5}}{\sqrt{LC}}$
we know that $W _{resonance}=\dfrac{1}{\sqrt{LC}} .... (2)$
Clearly from $(1)$ and $(2)$ $W>W _{res}(C)$
Multiple choice zener diode special purpose diodes electronic devices semiconductor electronics: materials, devices and simple circuits physics

 If the series resistance decreases in an unloaded zener regulator, the zener current

  1. Decreases

  2. Stays the same

  3. Increases

  4. Equals the voltage divided by the resistance
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

current increases as the resistance decreases, as zener current is inversely proportional to the series resistance