Answer: (a) 0.178 A
(b) 11.91 V
(c) 2.18 W / 0.016 W
Step-by-step explanation:
(a) To find the current I in the circuit, we can use Ohm's law:
V = IR,
where V is the voltage across the resistor R, I is the current in the circuit, and R is the resistance of the resistor. The voltage across the resistor is given by:
V = E - Ir,
where E is the emf of the battery, and r is the internal resistance of the battery. Substituting the given values, we get:
V = 12.0 V - 0.5 Ω x I,
65.0 Ω x I = 12.0 V - 0.5 Ω x I,
Solving for I, we get:
I = 0.178 A.
Therefore, the current in the circuit is 0.178 A.
(b) To find the terminal voltage of the battery, we can use the formula:
Vab = E - Ir.
Substituting the given values, we get:
Vab = 12.0 V - 0.5 Ω x 0.178 A = 11.91 V.
Therefore, the terminal voltage of the battery is 11.91 V.
(c) The power dissipated in the resistor R can be found using the formula:
P = I^2R.
Substituting the values we get:
P = (0.178 A)^2 x 65.0 Ω = 2.18 W.
The power dissipated in the battery's internal resistance can be found using the formula:
P = I^2r.
Substituting the values we get:
P = (0.178 A)^2 x 0.5 Ω = 0.016 W.
Therefore, the power dissipated in the resistor R is 2.18 W, and the power dissipated in the battery's internal resistance is 0.016 W.
This explanation takes about 2 hours to complete. Thank God it was done. Anyway, this is the explanation. Thank you.