Question
A 4.0 Ohm resistor it connected to the terminals of a non-ideal battery of emf 18 V. The current in the circuit is measured lo be 3.0 A Find the internal resistance of the battery Find the terminal potential difference of the battery In (c) through (e), a network consisting of two 60 Ohm resistors connected in parallel is placed m series with the 4.0 Ohm resistor Find the current in the circuit Find the potential difference across the circuit, i.e., the potential difference across the 4.0 Ohm resistor and the network of two 6.0 Ohm resistors Together Find the power output of the battery.
Explanation / Answer
(a) Given reistor : R1 = 4 emf of the battery = E = 18 volt current measured inthe circuit is : I = 3 A Let , r be the internal resistence of the battery It is known by the formula for internalresistence of the battrery r = E - I R / I = 18 - (3)(4)/(3) = 6 / 3 = 2 Thus, Internal resistence of the circuit is : r = 2 (b) Terminal potenaatil difference is : V = I R = (3)(4) = 12 Volt (c) since two 6 are connected in parallel effective resitence of them is : 6 x 6 / 6 + 6 R 2 = 3 R1 = 4 both are now connected int sereise thus, effective resistence is : R3 = 3 + 4 = 7 If inernal resistence also taken in to consideration , we have total resitence of the circiut is : R = 7 + r = 9 thus,current in the circuit is: I = E / R = 18 /9 = 2 A (d) Potenail difference across the 4 reiastor V4 = I R = (2) (4 ) = 8 Volt V6 = I R = (2) (3) = 6 volt (e) power in the circiut : P = V I = (18)(2) = 36 W