Question

Two resistors R_1 and R_2 are connected in series to a battery supplying a voltage of Delta V as shown in the figure. How much power is dissipated by the resistor R_2 in the form of heat? Use the following data: R_1 = 10 Ohm, R_2 = 5 Ohm, Delta V = 15 V. 2.38 3.45 5.00 7.25 1.05 times 10^1 1.52 times 10^1 2.21 times 10^1 3.20 times 10^1

Answer 1

Net resistance of circuit will be

$R=R_1+R_2=10+5=15$

Hence current will be

$I=\frac{V}{R}=\frac{15}{15}=1$

Hence power dissipated at $R_2$ will be

$H=i^2R_2=1^2\times5=5$ J$$