1

### JEE Main 2016 (Online) 9th April Morning Slot

In the circuit shown, the resistance r is a variable resistance. If for r = fR, the heat generation in r is maximum then the value of f is :
A
${1 \over 4}$
B
${1 \over 2}$
C
${3 \over 4}$
D
1

## Explanation

Here equivalent resistance

Req = R + ${{r \times R} \over {r + R}}$

=  R + ${{f{R^2}} \over {fR + R}}$

=   R + ${{fR} \over {f + 1}}$

=   ${{\left( {2f + 1} \right)R} \over {\left( {f + 1} \right)}}$

Circuit current,

${\rm I}$ = ${V \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$

=   ${{V\left( {f + 1} \right)} \over {{\mathop{\rm R}\nolimits} \left( {2f + 1} \right)}}$

Current in the resistance r,

I = ${{\rm I} \over {f + 1}}$ = ${V \over {R\left( {2f + 1} \right)}}$

Heat generated in r,

H = ${\rm I}_2^2\,r$

=  ${{{V^2}f} \over {R{{\left( {2f + 1} \right)}^2}}}$

For maximum H,

${{dH} \over {df}}$ = 0

$\Rightarrow$   ${{{V^2}} \over R}\left[ {{1 \over {{{\left( {2f + 1} \right)}^2}}} - {{4f} \over {{{\left( {2f + 1} \right)}^3}}}} \right]$ = 0

$\Rightarrow$   2f + 1 = 4f

$\Rightarrow$   2f = 1

$\Rightarrow$   f = ${1 \over 2}$
2

### JEE Main 2016 (Online) 10th April Morning Slot

The resistance of an electrical toaster has a temperature dependence given by R(T) = R0 [1 + $\alpha$(T − T0)] in its range of operation. At T0 = 300 K, R = 100 $\Omega$ and at T = 500 K, R = 120 $\Omega$. The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate from 300 to 500 K in 30 s. The total work done in raising the temperature is :
A
400 $\ln \,{{1.5} \over {1.3}}\,J$
B
200 $\ln \,{{2} \over {3}}\,J$
C
60000 $\ln \,{{6} \over {5}}\,J$
D
300 J

## Explanation

Given,

R = R6 [1 + $\alpha$ (T $-$ t0)]

120 = 100 [1 + $\alpha$ (500 $-$ 300)]

$\Rightarrow$   200 $\alpha$ = ${1 \over 5}$

$\Rightarrow$   $\alpha$ = 10$-$3   oC$-$1

Temperature of the toaster raised from 300 K to 500 K in 30 s.

$\therefore$   Increment in the temperature in time t,

$\Delta$T = ${{500 - 300} \over {30}}t$

= ${{200} \over 3}t$

= ${{20} \over 3}t$

Total work done in raising the temperature

= $\int\limits_0^t {{{{V^2}} \over {R\left( t \right)}}\,dt}$

= $\int\limits_0^t {{{{V^2}} \over {{R_0}\left( {1 + \alpha \Delta t} \right)}}} \,dt$

= $\int\limits_0^{30} {{{{{\left( {200} \right)}^2}} \over {100\left( {1 + {{10}^{ - 3}} \times {{20} \over 3}t} \right)}}} \,dt$

= ${{40000} \over {100}}\int\limits_0^{30} {{{dt} \over {\left( {1 + {t \over {150}}} \right)}}}$

$400 \times 150\left[ {\ln \left( {1 + {t \over {150}}} \right)} \right]_0^{30}$

$= 60000\left[ {\ln \left( {1 + {{30} \over {150}}} \right) - \ln 1} \right]$

$= 60000\ln \left( {{6 \over 5}} \right)\,J$
3

### JEE Main 2017 (Offline)

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be:
A
$CE{{{r_1}} \over {({r_1} + r)}}$
B
CE
C
$CE{{{r_1}} \over {({r_2} + r)}}$
D
$CE{{{r_2}} \over {(r + {r_2})}}$

## Explanation

In steady state, flow fo current through capacitor will be zero.

Current through the circuit,

i = ${E \over {r + {r_2}}}$

Potential difference through capacitor

Vc = ${Q \over C}$ = E - ir = E - $\left( {{E \over {r + {r_2}}}} \right)r$

$\therefore$ Q = $CE{{{r_2}} \over {(r + {r_2})}}$
4

### JEE Main 2017 (Offline)

In the given circuit, the current in each resistance is:
A
0 A
B
1 A
C
0.25 A
D
0.5 A

## Explanation

The potential difference in each loop is zero.

$\therefore$ No current will flow or current in each resistance is Zero.