A) 0.5 V
B) 1.9 V
C) 1.95 V
D) 2 V
Correct Answer: C
Solution :
| [c] Key Idea: Terminal voltage across the cell decreases due to voltage drop across internal resistance. |
| The current flowing in the circuit is |
| \[i=\frac{E}{R+r}\] |
| Given, \[E=2\text{ }V,\text{ }R=3.9\,\Omega ,\] |
| \[r=0.1\,\Omega \] |
| So, \[i=\frac{2}{3.9+0.1}=\frac{2}{4.0}=0.5\,A\] |
| The voltage drop across internal resistance, |
| \[V'=ir=0.5\times 0.1\] |
| \[1=0.05\text{ }V\] |
| Thus, terminal voltage across cell is, |
| \[V=E-ir=E-V'\] |
| \[=2-0.05=1.95\,V\] |
You need to login to perform this action.
You will be redirected in
3 sec
