A cell of emf E and internal resistance r supplies currents for the same time t through external resistance R1= 100 Ω and R2=40 Ω separately. If the heat developed in both cases is the same, then the internal resistance of the cell is given by

Question

A cell of emf E and internal resistance r supplies currents for the same time t through external resistance R1= 100 Ω and R2=40 Ω separately. If the heat developed in both cases is the same, then the internal resistance of the cell is given by (A) 28.6 Ω  (B) 70 Ω  (C) 63.3 Ω (D) 140 Ω

Tardigrade Admin · Accepted Answer

Current drawn from the cell in resistance

R_{1}

will be

I = E (R_{1} + r)

Therefore, the heat produced in

R_{1}

i.e.

H_{1} = \frac{E ^{2} R _{1} t}{( R _{1} + r ) ^{2}}

Heat produced in

R_{2}, H_{2} = \frac{E ^{2} R _{2} t}{( R _{2} + r ) ^{2}}

As per question

H_{1} = H_{2}

or

\frac{E ^{2} R _{1} t}{( R _{1} + r ) ^{2}} =

\frac{E ^{2} R _{2} t}{( R _{2} + r ) ^{2}}

On solving we get;

r = R_{1} R_{2}

= 100 \times 40 = 63.25 Ω

Q. A cell of emf $E$ and internal resistance $r$ supplies currents for the same time $t$ through external resistance $R_{1} = 100 Ω$ and $R_{2} = 40 Ω$ separately. If the heat developed in both cases is the same, then the internal resistance of the cell is given by

$28.6 Ω$

$70 Ω$

$63.3 Ω$

$140 Ω$

Q. A cell of emf E and internal resistance r supplies currents for the same time t through external resistance R1​=100Ω and R2​=40Ω separately. If the heat developed in both cases is the same, then the internal resistance of the cell is given by

28.6Ω

70Ω

63.3Ω

140Ω

Q. A cell of emf $E$ and internal resistance $r$ supplies currents for the same time $t$ through external resistance $R_{1} = 100 Ω$ and $R_{2} = 40 Ω$ separately. If the heat developed in both cases is the same, then the internal resistance of the cell is given by

$28.6 Ω$

$70 Ω$

$63.3 Ω$

$140 Ω$