Question

In the circuit shown in the figure the heat produced in the 5ohm resistor due to the current flowing through it is 10 calories per second. The heat generated in the 4 ohm resistor is
A. 1calorie/per
B. 2 calorie/per
C. 3 calorie/per
D. 4 calorie/per

Answer 1

In such a type of question we must know the concept of resistor in series and parallel and the heating effect of electric current and how I and V vary in series and parallel combination.

Formula used - $I_4^2{R_4} = {(\dfrac{{\sqrt 2 }}{2})^2} \times 4 = 2cal/\sec$

1. $H = {I^2}Rt$
2. $V = IR$
3. Current through a resistor in series combination, ${I_S} = \dfrac{V}{{{R_1} + {R_2}}}$

Complete step-by-step answer:
Any resistor in a circuit that has a voltage drop over it dissipates electrical power. This electrical power is converted to heat energy, and thus all resistors have a power rating. The conversion rate is the dissipation power.
Now, it is given that,
Heat generated at 5 ohms resistor=10 calories per second
Heat yielded, $H = {I^2}Rt$
Here, $\dfrac{{{H_5}}}{t} = {I_5}^2{R_5} = 10 \Rightarrow {I_5} = \sqrt {\dfrac{{10}}{5}} = \sqrt 2$, This is the current through the resistor to 5 ohm.

Voltage through 5 ohms resistor ${V_5} = {I_5}{R_5} = \sqrt 2 \times 5 = 5\sqrt 2 V$
As the resistors (4,6) and 5 are parallel, the overall potential (4,6) is equal to that of ${V_5}$
As the 4 and 6 resistors are so present through each resistor in series ${I_4} = {I_6} = \dfrac{{{V_5}}}{{4 + 6}} = \dfrac{{5\sqrt 2 }}{{4 + 6}} = \dfrac{{\sqrt 2 }}{2}$
Heat produce in resistor 4 per sec,$I_4^2{R_4} = {(\dfrac{{\sqrt 2 }}{2})^2} \times 4 = 2cal/\sec$
Hence the correct answer is option B.

Note: In such a type of question, we must apply the formula on the given resistor to obtain V and I in the resistor given and then apply the concept and formula of series and parallel combination for finding the required quantity.