Question

A battery of e.m.f. 15 V and internal resistance 2 $\Omega$ is connected to two resistors of resistances 4 $\Omega$ and 6 $\Omega$joined in series. Find the electrical energy spent per minute in 6 $\Omega$ resistor.

Answer 1

Given,
E.m.f.of battery , $v=15v$
Internal resistance of battery,$R_B=2\Omega$
Resistance given in circuit,$R_1=4\Omega$
$R_2=6\Omega$
i) When resistors re connected in series
Equivalent resistance,$R=R_B+R_1+R_1+R_2=12\Omega$
Current the circuit,$I=\frac{15}{12}=1.25A$
Now voltage across resistors $R_2,V_2=IR=1.25\times 6$
$V_2=7.50V$
Time,t=1 min=60sec
Energy across $R_2,E=\frac{V^{2}t}{R}=\frac{(7.5{)}^2\times 60}{6}$
$E=562.5J$