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

An ac signal of 1 V (rms) is applied to the grid of a triode for which (mu) u = 10 and plate resistance Rp = 15 k ohm. The signal frequency is low enough for the interelectrode capacitance to be negligible. The load impedance ZL = (2 + j 10) k ohm. Calculate voltage gain of the amplifier.

  1. 0.404

  2. 4.047

  3. 8.231

  4. 1.500

  5. -
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Plate resistance (Rp) = 15 kilo ohm Amplification factor (mu) u = 10 Load impedence (ZL) = (2 + j10) kilo ohm

Multiple choice

A strain gauge has a resistance of 120 ohm and gauge factor of 2. This strain gauge is connected in series with a ballast resistance of 120 ohm across a 6 V supply. Determine the voltage sensitivity of the circuit when the gauge is subjected to pressure of 240 MN/m2. The modulus of elasticity is 200 GN/m2.

  1. 5 volts per unit strain

  2. 48 volts per unit strain

  3. 1.98 volts per unit strain

  4. 3 volts per unit strain
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Ballast resistance (Rb) = 120 ohm Strain gauge resistance (Rs) = 120 ohm Input voltage (Vi) = 6 V Gauge Factor = (G) = 2 Applied Stress (p)= 240 MN/ m2 Modulus of elasticity =(E) = 200 GN/m2 Strain= P/E(240 x 106)/(200 x 109) = 1.2 x 10-3 Voltage sensitivity of the circuit (sV) = ((Rb x Rs.)/(Rb+ Rs.)2) x G x Vi (120 x 120)/(120+120)2) x 2 x 6 = 3 volts per unit strain 

Multiple choice

A linear resistance potentiometer is 50 mm long and is uniformly wound with a wire of total resistance 5000 ohm. Under normal conditions, the slider is at the centre of the potentiometer. Determine linear displacement when the resistance of the pot, as measured by the Wheatstone bridge is 1850 ohm. If it is possible to measure the value of 5 ohm resistance with the above arrangement, determine the resolution of pot in mm.

  1. 4.5 mm and 0.5 mm

  2. 6.5 mm and 0.05 mm

  3. 65 km and 0.05 mm

  4. 6.5 mm and 0.005 mm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total resistance of the pot wire (Rt) = 5000 ohm Total length of the pot wire (L) = 50 mm Resistance per mm length of pot wire (Rt/L) = 5000/50 = 100 ohm Resistance of pot at normal position = 5000/2 = 2500 ohm Change in resistance of pot from its normal position = 2500 – 1850 = 650 ohm Dispalcement (y) = 650/100 = 6.5 mm Minimum measurable value of resistance = 5 ohm Resolution of pot = Minimum measurable resistance/Resistance per mm length = 5/100 = 0.05 mm 6.5 mm and 0.05 mm 

Multiple choice

How can an ammeter be converted into a galvanometer?

  1. Connecting a very high resistance in parallel

  2. Connecting a very high resistance in series

  3. Connecting a shunt in series

  4. Connecting a shunt parallel to it

  5. Cannot be converted
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

It will become an ammeter.

Multiple choice physics electric current and its effects coil and fuse heating effect of electric current- uses applications of heating effect of electric current

Current through resistor of resistance $R$ varies with time as $I = I _ot$. What will be the time taken to deliver heat $H$ by this resistor?

  1. $\left( \cfrac { H } { I _ { 0 } R } \right) ^ { 1 / 2 }$
  2. $\left( \cfrac { H } { 3 I _ { 0 } ^ { 2 } R } \right) ^ { 1 / 3 }$
  3. $\left( \cfrac { 3 H } {I _ { 0 } ^ { 2 } R } \right) ^ { 1 / 3 }$
  4. $\left( \cfrac { H } { I _ { 0 } ^ { 2 } R } \right) ^ { 1 / 3 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Heat produced is H = ∫I²R dt. Given I = I₀t, we have H = ∫(I₀t)²R dt from 0 to T. This gives H = I₀²R∫t²dt = I₀²R(T³/3). Solving for time T: T³ = 3H/(I₀²R), so T = (3H/(I₀²R))^(1/3). This matches option C.

Multiple choice physics electric current and its effects coil and fuse heating effect of electric current- uses applications of heating effect of electric current

Two equal resistance when connected in series to a battery, consume electri power  of 60 w.  If these resistnce are now  connected in parallel combination to  the same battery the electric power consumed will be 

  1. 60 W

  2. 240 W

  3. 30 W

  4. 120 W
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When two equal resistors of resistance R are in series, the equivalent resistance is 2R and power is V^2 / (2R) = 60 W. When connected in parallel, the equivalent resistance is R/2, so the power becomes V^2 / (R/2) = 4 * (V^2 / 2R) = 4 * 60 W = 240 W.

Multiple choice physics current electricity and magnetism electric cells and batteries

While making a battery ___________  of a cell is connected to ________ of other cell.

  1. negative terminal , negative terminal

  2. positive terminal , negative terminal

  3. positive terminal , positive terminal

  4. neutral terminal ,negative terminal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

While making a battery, negative end of battery is connected to positive terminal of other cell and so on.


Answer-(B)

Multiple choice physics chemical effects of electric current study about cells electric cell types of cells

An inductor coil stores $U$ when a current is passed through it and dissipates heat energy at the rate of $P .$ The time constant of the circuit when this connected across a battery of zero internal resistance is:


  1. $\frac { 4 U } { P }$
  2. $\frac { U } { P }$
  3. $\frac { 2 U } { P }$
  4. $\frac { 2 P } { 6U }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Energy stored U = (1/2)LI^2. Power dissipated P = I^2R. Time constant tau = L/R. From equations, L = 2U/I^2 and R = P/I^2. Thus, tau = (2U/I^2) / (P/I^2) = 2U/P.

Multiple choice physics chemical effects of electric current study about cells electric cell types of cells

Five cells each of emf $\varepsilon $ and internal resistance r each are connected in series. If one cell is connected wrongly, then potential difference across this cell will be

  1. $\dfrac { 8\varepsilon }{ 5 } $

  2. zero
  3. $\dfrac { 4\varepsilon }{ 5 } $
  4. $\dfrac { 2\varepsilon }{ 5 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When five cells of emf epsilon and internal resistance r are in series, the total emf is 5 epsilon and total internal resistance is 5r. If one cell is connected with reverse polarity, the net emf becomes 5 epsilon - 2 epsilon = 3 epsilon, and the current in the circuit is I = (3 epsilon) / (5r). The potential difference across the wrongly connected cell is V = epsilon_cell + I*r = epsilon + ((3 epsilon)/(5r))*r = epsilon + (3/5)epsilon = (8 epsilon) / 5.

Multiple choice physics chemical effects of electric current study about cells electric cell types of cells

A wire of length $L$ and $3$ identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by $\Delta\ T$ in a time $t$. A number $N$ of similar cells is now connected in series with a wire of the same material and cross-section but of length $2\ L$. The temperature of the wire is raised by the same amount $\Delta\ T$ in the same time. The value of $N$ is

  1. $4$
  2. $6$
  3. $8$
  4. $9$
Reveal answer Fill a bubble to check yourself
B Correct answer