Two resistors `R_1 and R_2` of resistance 3 `Omega ` and 6 `Omega `respectively are connected in parallel across a battery of p.d. 12 V. Draw the circuit diagram. Calculate the electrical energy consumed in 1 minute in each resistance.

Explanation

The circuit is shown in Fig

Given : `R_1 = 3 Omega , R_2 = 6 Omega , V = 12` volt,
` t=1` minute `= 60 s`
Total resistance of circuit `R = R_1 + R_2`
`= 3+6 = 9 Omega `
current in circuit `I=(V )/(R ) = (12)/(9) = 4/3 A`
since the resisitance are in series , so same current flows in each resistance .
Electrical energy consumed in` R_1 ` will be
` W_1 = I2 R_1 t = ((4)/(3))^2 xx 3 xx 60 = 320 J`
Electrical energy consumed in `R_2` will be
` W_2 =I^2 R_2 t = (4/3)^2 xx 6 xx 60 = 640 J`