91影视

a) Resistance of the resistor,$R=0.10\text{鈥�}惟$ .

b) Rate of energy transferred as thermal energy,$P=10\text{鈥塛}$ ,

c) Emf of the battery,$蔚=1.5\text{鈥塚}$ .

Here, the rate of energy transferred to the resistor as thermal energy is given. Thus, using the concept, the value of the external resistance can be given. Now, the current through the battery and the resistor is equal to the closed loop rule. Thus, equating the current values will determine the internal resistance of the battery.

Formulae:

The rate at which the thermal energy is transferred,

$P=\frac{V^2}{R}$ (i)

The voltage equation using Ohm鈥檚 law,

$V=IR$ (ii)

Here $I$is the current, and $R$is the resistance.

The potential difference across the resistor due to the rate of energy transferred as thermal energy can be given using equation (i) as follows:

$V=\sqrt{PR}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}V=\sqrt{\left(10\text{鈥塛}\right)\left(0.10\text{鈥�}惟\right)}\\ =1.0\text{鈥塚}\end{array}$

Hence, the value of the potential difference is.$1.0\text{鈥塚}$

Now, we know that current through the resistor and the battery is equal, and thus, the internal resistance of the battery can be given using equation (ii) as follows:

$\begin{array}{c}{i}_R=i_{battery}\\ \frac{V}{R}=\frac{蔚鈭�V}{r}\\ r=\left(\frac{蔚鈭�V}{V}\right)R\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}r=\left(\frac{1.5\text{鈥塚}鈭�1.0\text{鈥塚}}{1.0\text{鈥塚}}\right)0.10\text{鈥�}惟\\ =0.05\text{鈥�}惟\end{array}$

Hence, the value of internal resistance is.$0.05\text{鈥�}惟$