A battery of emf 15V and internal resistance 2ohm is connected to two resistors of resistances 4ohm and 6ohm joined in series. Find the electrical energy spent per minute in 6ohm resistor.
Answers
r=2ohm
R=4+6=10ohm
I=Emf/r+R
after solving we get I=15/12A
electrical energy spent is equal to power X 60=I^2R =(1.25)^2 X 6X60= 562.5J
here's the answer.
Answer:
669.42 J
Explanation:
When the resistors are joined in series, the total resistance of the circuit is 4 + 6 + 2 = 12 ohm. The current passing through all the resistors is I = V/R = 15/12.
The current passing through the 6 ohm resistor is 1.25 A. The energy spent due to this is equal to power*60 = (1.25)^2*6*60 = 562.5 J
b. When the resistors are connected in parallel, their total resistance is (1/4 + 1/6)^-1 = 2.4 ohm. With the internal resistance, the total resistance of the circuit is 2 + 2.4 = 4.4 ohm. A voltage of 15 V causes a current of 15/4.4 = 75/22 A to flow through the circuit. This leads to a voltage of 90/11 V across the two resistors. The current through the 6 ohm resistor is 15/11 A.
The energy spent due to this in a minute is (15/11)^2*6*60 = 669.42 J.
HOPE IT HELPED :)