JEE MAIN - Physics (2015 (Offline) - No. 11)
_en_11_1.png)
In the circuit shown, the current in the $$1\Omega $$ resistor is :
$$0.13$$ $$A,$$ from $$Q$$ to $$P$$
$$0.13$$ $$A$$, from $$P$$ to $$Q$$
$$1.3A$$ from $$P$$ to $$Q$$
$$0A$$
Explanation
From $$KVL$$
$$ - 6 + 3{{\rm I}_1} + {\rm 1}\left( {{{\rm I}_1} - {{\rm I}_2}} \right) = 0$$
_en_11_2.png)
$$6 = 3{{\rm I}_1} + {{\rm I}_1} - {{\rm I}_2}$$
$$4{{\rm I}_1} - {{\rm I}_2} = 6\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
$$ - 9 + 2{{\rm I}_2} - \left( {{{\rm I}_1} - {{\rm I}_2}} \right) + 3{{\rm I}_2} = 0$$
$$ - {{\rm I}_1} + 6{{\rm I}_2} = 9\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
On solving $$\left( 1 \right)$$ and $$\left( 2 \right)$$
$${{\rm I}_1} = 0.13A$$
Direction $$Q$$ to $$P,$$ since $${{\rm I}_1} > {{\rm I}_2}.$$
Alternatively
_en_11_3.png)
$$Eq = {{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}}}}$$
$$ = {{{6 \over 3} - {9 \over 5}} \over {{1 \over 3} + {1 \over 5}}} = {3 \over {8V}}$$
$$\therefore$$ $${\rm I} = {{{3 \over 8}} \over {{{15} \over 8} + 1}} = {3 \over {23}} = 0.13A$$
Considering potential at $$P$$ as $$0V$$ and at $$Q$$ as $$x$$ volt, then
_en_11_4.png)
$${{x - 6} \over 3} + {{x - 0} \over 1} + {{x + 9} \over 5} = 0$$
$$\therefore$$ $$x = {2 \over {23}}$$
$$\therefore$$ $$i = {{x - 0} \over 1} = {2 \over {23}} = 0.13A$$
From $$Q$$ to $$P$$
$$ - 6 + 3{{\rm I}_1} + {\rm 1}\left( {{{\rm I}_1} - {{\rm I}_2}} \right) = 0$$
$$6 = 3{{\rm I}_1} + {{\rm I}_1} - {{\rm I}_2}$$
$$4{{\rm I}_1} - {{\rm I}_2} = 6\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
$$ - 9 + 2{{\rm I}_2} - \left( {{{\rm I}_1} - {{\rm I}_2}} \right) + 3{{\rm I}_2} = 0$$
$$ - {{\rm I}_1} + 6{{\rm I}_2} = 9\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
On solving $$\left( 1 \right)$$ and $$\left( 2 \right)$$
$${{\rm I}_1} = 0.13A$$
Direction $$Q$$ to $$P,$$ since $${{\rm I}_1} > {{\rm I}_2}.$$
Alternatively
$$Eq = {{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}}}}$$
$$ = {{{6 \over 3} - {9 \over 5}} \over {{1 \over 3} + {1 \over 5}}} = {3 \over {8V}}$$
$$\therefore$$ $${\rm I} = {{{3 \over 8}} \over {{{15} \over 8} + 1}} = {3 \over {23}} = 0.13A$$
Considering potential at $$P$$ as $$0V$$ and at $$Q$$ as $$x$$ volt, then
$${{x - 6} \over 3} + {{x - 0} \over 1} + {{x + 9} \over 5} = 0$$
$$\therefore$$ $$x = {2 \over {23}}$$
$$\therefore$$ $$i = {{x - 0} \over 1} = {2 \over {23}} = 0.13A$$
From $$Q$$ to $$P$$
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