Q 1 :    

The terminal voltage of the battery, whose emf is 10 V and internal resistance 1Ω, when connected through an external resistance of 4Ω as shown in the figure is    [2024]

[IMAGE 185]----------------------------------

  • 4 V

     

  • 6 V

     

  • 8 V

     

  • 10 V

     

(3)

Current in the given circuit

I=104+1=105=2A

Terminal voltage, V=E-Ir=10-2×1=8V



Q 2 :    

10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased n times. The value of n is                    [2023]

  • 1

     

  • 1000

     

  • 10

     

  • 100

     

(4)

Net resistance in series combination,

RN=R1+R2++R10

=10+10++10=10×10=100Ω                      ...(i)

Same number of resistors and each of same magnitude connected in parallel combination,

      R'N=(1R1+1R2++1R10)-1

            =(1+1++110)-1=1010=1Ω                               ...(ii)

Using (i) and (ii), I'I=V.RNR'N.V=1001

Thus, option (4) is correct.



Q 3 :    

A set of n equal resistors, of value R each, are connected in series to a battery of emf E and internal resistance R. The current drawn is I. Now, the n resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10I. The value of n is             [2018]

  • 10

     

  • 11

     

  • 20

     

  • 9

     

(1)

Current drawn from a battery when n resistors are connected in series is

         I=EnR+R                         ...(i)

Current drawn from same battery when n resistors are connected in parallel is

       10I=ERn+R                       ...(ii)

On dividing eqn. (ii) by (i), 10=(n+1)R(1n+1)R

After solving the equation, n=10