Physics

Current Electricity and Circuits

450 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

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

In the measurement of resistance by a metre bridge, the known and unknown resistance are interchanged to eliminate 

  1. end error

  2. index error

  3. random error

  4. error due to thermoelectric effect
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

End errors in a meter bridge arise due to end resistances at the copper strips and resistance of the connecting wires at the zero and hundred centimeter marks. Interchanging the known and unknown resistances helps cancel out these systematic end errors by taking the average of the two balancing lengths.

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

A $6\Omega$ resistance is connected in the left gap of a meter bridge. In the second gap $3\Omega$ and $6\Omega$ are joined in parallel. The balance point of the bridge is at __

  1. $75cm$
  2. $60cm$
  3. $30cm$
  4. $25cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the right gap, a 3 ohm and 6 ohm resistor are in parallel, giving an equivalent resistance of (3 * 6) / (3 + 6) = 18 / 9 = 2 ohms. Let the balancing length from the left be l. For a meter bridge, R1 / R2 = l / (100 - l), so 6 / 2 = l / (100 - l), which gives 3 = l / (100 - l). Solving this yields 300 - 3l = l, so 4l = 300, giving l = 75 cm.

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

A resistor $ ^{\prime} R^{\prime}  $ and $2  \mu F  $ capacitor in series is connected through a switch to $200  \mathrm{V}  $ direct supply. Across the capacitor is a neon bulb that lights up at $120  \mathrm{V} $ Calculate the value of $  R  $ to make the bulb light up $5  s  $ after the switch has been closed. $ \left(\log _{10} 2.5=0.4\right) $

  1. $2.7 \quad 10^{6} \Omega $
  2. $3.3 \quad 10^{7} \Omega $
  3. $1.3 \quad 10^{4} \Omega $
  4. $1.7 \quad 10^{5} \Omega $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The voltage across a charging capacitor is Vc = V0 * (1 - exp(-t/RC)). Given Vc = 120, V0 = 200, t = 5, and C = 2 * 10^-6, we have 120 = 200 * (1 - exp(-5/(R * 2 * 10^-6))). Simplifying gives 0.6 = 1 - exp(-5/(2 * 10^-6 * R)), so 0.4 = exp(-5/(2 * 10^-6 * R)). Taking the natural log, ln(0.4) = -5/(2 * 10^-6 * R). Using log10(2.5) = 0.4, we find R is approximately 2.7 * 10^6 ohms.

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 physics electric current through conductors carbon resistance and colour codes for carbon resistance carbon resistors and their colour coding resistors

Which of the following is not true for wire wound resistor?

  1. It has a lower order of stability and reliability.

  2. It has high power rating with a low tolerance value.

  3. Easy to make wire wound resistor of value 0.01 Ohm.

  4. It is not suitable for high-frequency circuits.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
A wire wound resistor is an electrical passive component that limits current. The resistive element exists out of an insulated metallic wire that is winded around a core of non-conductive material. The wire material has a high resistivity, and is usually made of an alloy such as Nickel-chromium (Nichrome) or a copper-nickel-manganese alloy called Manganin. Common core materials include ceramic, plastic and glass. 
It has a lower order of stability and reliability  is not true for wire wound resistor
Multiple choice physics electric current through conductors carbon resistance and colour codes for carbon resistance carbon resistors and their colour coding resistors

Which of the following is not the disadvantage of wire wound resistor?

  1. It has a big size which is not suitable for many of the applications.

  2. It has high power rating with a low tolerance value.

  3. Resistance value gets changed by a change in temperature, humidity etc.

  4. It is not suitable for high-frequency circuits.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In wired resistors are mainly produced with alloys, since pure metal has a high temperature co-efficient.

Due to alloy temperature co-efficient of wire wound resistor is very low.
Therefore it is very less effected by temperature change.