QJEEM15Ph28e7: JEE Main 2015 Physics - Class 12 JEEM15Ph28e7 Solution: In the circuit shown, the current in the resistor is :

Question

Verified Solution • 2025–26 Session moderate

In the circuit shown, the current in the

1\Omega

resistor is :

Accepted Answer

From

KVL

- 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} &gt; {{\rm I}_2}.

Alternatively

JEE Main 2015 (Offline) Physics - Current Electricity Question 297 English Explanation 2

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

JEE Main 2015 (Offline) Physics - Current Electricity Question 297 English Explanation 3

{{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

Step-by-Step Verified Solution