| A 5V battery with internal resistance \[2\,\Omega \] and a 2V battery with internal resistance \[1\,\Omega \] are connected to a \[10\,\,\Omega \] resistor as shown in the figure. The current in the \[10\,\,\Omega \] resistor is [AIEEE 2008] |
 |
A) 0.03 A \[{{P}_{2}}\] to \[{{P}_{1}}\]
B) 0.27 A \[{{P}_{1}}\] to \[{{P}_{2}}\]
C) 0.27 A \[{{P}_{2}}\] to \[{{P}_{1}}\]
D) 0.03 A \[{{P}_{1}}\] to \[{{P}_{2}}\]
Correct Answer: A
Solution :
[a] \[i=\frac{{{\varepsilon }_{1}}{{r}_{2}}+{{\varepsilon }_{2}}{{r}_{1}}}{{{r}_{1}}{{r}_{2}}+R{{r}_{1}}+R{{r}_{2}}}=\frac{5\times 1+\left( -2 \right)\times 2}{2\times 1+10\times 2+10\times 1}\] = 0.0.3 A
You need to login to perform this action.
You will be redirected in
3 sec
