Current Electricity - Online Test

Q1. A Wheatstone bridge ABCD is balanced with a galvanometer between the points B and D. At balance the resistance between the points B and D is :
Answer : Option D
Explaination / Solution:

No current passes through the Galvanometer in the bridge balance condition. Therefore, the resistance between B and D is infinite.

Q2. The ratio of the currents I1 when switch S is off and I2 when switch S is on is

Answer : Option B
Explaination / Solution:

When swithch S is off, both resistances are included in the circuit.



Q3. An electric bulb rated for 500 watts at 100 volts is used in a circuit having 200 volt supply. The resistance R that must be put in series with the bulb so that the bulb draws 500 watts is
Answer : Option C
Explaination / Solution:

The resistance of the bulb RB = V2/P = 1002/500 = 20\Omega 

When connected across a 200V supply with a resistor R in series, the power drawn remains the same. If V1 be the p.d across the bulb,V12 = P {RB} = 500 * 20 = 104;

V1 = 100V, . Therefore the p.d across R =200-100=100 V. Therefore R=RB=20Ω


Q4. In the circuit shown below, the galvanometer G will show zero deflection

Answer : Option C
Explaination / Solution:

The ratio of the resistances in the arms is not equal. The wheatstone’s network is not balanced. The galvanometer will continue to show deflection in all cases.

Q5. A milliammeter of range 10 mA has a coil of resistance 1 Ω.To use it as an ammeter of range 1 A, the required shunt must have a resistance of
Answer : Option C
Explaination / Solution:



(Theory in chapter 4 – magnetic effect of current)

Q6. A capacitor is charged and then made to discharge through a resistance. The time constant is τ. In what time will the potential difference across the capacitor decreases by 10%?
Answer : Option D
Explaination / Solution:

The capacitance of a capacitor is a constant.

 

Also, V  q

The discharging equation for capacitor is 

 

A similar equation can be written for potential difference.


Q7. Meter Bridge is used to
Answer : Option A
Explaination / Solution:

With a known resistance in one of the gaps, the meter bridge is used to determine the value of an unknown resistance by the formula, 


where l is the null point.


Q8. Resistance of a conductor is
Answer : Option B
Explaination / Solution:

According to Ohm’s law, V=I R. Therefore R=V/I

Q9.

For the given circuit , with cell e m f = 10 V and internal resistance = , which of the following is correct?

Answer : Option C
Explaination / Solution:

The circuit is reduced to find equivalent resistance as follows;





In (1), 2Ω, 4 Ω, and 2 Ω are in series. Their equivalent resistance is 8 Ω. In (2), the two 8 Ω resistors are in parallel and their equivalent resistance is 4 Ω. In (3) 2 Ω, 4 Ω, and 2 Ω are in series. Their equivalent resistance is 8 Ω which is in parallel with the 8 Ω resistance as shown in (4). The total resistance in the circuit

R= 3+4+2+1( internal resistance)=10 Ω. The current through the 3 Ω resistor

I = E/R = 10/10

I = 1A



Q10. Three voltmeters, all having different resistances, are joined as shown. When some potential difference is applied across A and B, their readings are V1, V2 , V3 .

Answer : Option D
Explaination / Solution:

V3 and combination of V1 and V2 (V1 + V2) are connected in parallel so potential in V3 must be equal to potential in (V1 + V2)

the potential difference across AB =  = +.