Question

The internal resistance of a cell of emf 4 V is $ 0.1 \Omega $ It is connected to a resistance of $ 3.9 \Omega $. The voltage across the cell will be

• A. 3.9V
• B. 2V
• C. 0.1V
• D. 3.8V

Answer 1

Answer: (A)

emf of a cell (E)=4 V internal resistance of a cell $ (r) = 0.1 \Omega $ external resistance $ (R) = 3.9 \Omega $
The potential drop across the cell $ \, \, \, \, \, V = E - 1.r \, \, \, \, \, ......(i) $
Now, the total resistance of the circuit
$ \, \, \, \, \, R' = r + R $
$ \, \, \, \, \, \, R' = 0.1 + 3.9 $
$ \Rightarrow \, \, \, \, R' = 4.0 \Omega$
Hence, current in the circuit is $ \, \, \, \, \, \, I = \frac{E}{R'} $
$ \Rightarrow \, \, \, \, I = \frac{4}{4} = 1 A $
Now, form Eq. (i)
$ V = 4 - 1 \times 0.1 $
$ V = 4 - 0.1 $
$ V = 3.9 volt $