A 12 Volt battery is connected in a circuit with a 1 ohm resistor and a 5 ohm resistor. They are in parallel and the 1 ohm is an internal resistance. Find the rate of conversion of internal (chemical) energy to electrical energy within the battery. Find the rate of dissipation of electrical energy in the battery. Find the rate of dissipation of electrical energy in the external resistor. A 12 Volt battery is connected in a circuit with a 1 ohm resistor and a 5 ohm resistor. They are in parallel and the 1 ohm is an internal resistance. Find the rate of conversion of internal (chemical) energy to electrical energy within the battery. Find the rate of dissipation of electrical energy in the battery. Find the rate of dissipation of electrical energy in the external resistor.
Explanation / Answer
A 12.0 V battery is connected in a circuit emf e = 12.0 V Internal resistance r = 1.0 ohms , 5.0 ohms V = e - I r I R = e - I r I ( R + r ) = e I = e / ( R + r ) R = 5.0 ohms * 1.0 ohms / 5.0 ohms * 1.0 ohms = 0.83 ohms I = 12.0 V / ( 0.83 + 1 ) = 6.54 A a ) P = I ^2 r = ( 6.54 A )^2 * 1.0 ohms = 42.8 W b ) P = e I = 12.0 V * 6.54 A = 78.4 W c ) P = I ^2 R = ( 6.54 A ) ^2 * 0.83 = 35.5 W