A cell has an e.m.f of 1.08 v and an internal resistor of 0.5 ohm. when it is connected in series with resistor of R , the potential difference between the terminals fell to 0.96 v . what was the value of R?

1 Answer
Ashley H · Stefan V.
Aug 13, 2018

#4 Omega#

Explanation:

The circuit diagram in the given case appears as follows,

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Here, you can place the internal resistance as a separate resistor in the circuit and in that case potential drop across the cell will be #V_A - V_B#

Now, it is given that the potential drop across the cell is #0.96V#; that means this drop is due to some potential drop across the #0.5 Omega# internal resistance in the cell, we can say the rest i.e #(1.08-0.96)=0.12V# has dropped across #0.5 Omega#.

So, if the current flowing through the circuit is #I#, then we can say, #I×0.5=0.12#

Or, #I=0.24 A#

Now, the voltage drop across resistor #R# is #(1.08-0.12)=0.96V# (note it is the same as #V_A-V_B#)

So,if voltage drop across #R# is #0.96 V# and Current flowing is #0.24A#

Then, using Ohm's Law, we can write,

#R=0.96/0.24=4 Omega#

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