Consider a circuit with a battery of electromotive force (emf) 10V and internal resistance 2Ω connected to a resistor of 4Ω. The circuit is closed and a current of 1A is passing through the circuit.
a) Calculate the potential difference across the resistor using Ohm's law. b) Calculate the power dissipated by the resistor. c) If the internal resistance of the battery is increased to 4Ω, what will be the new potential difference across the resistor?
Ohm's law states that the potential difference (V) across a resistor is directly proportional to the current (I) passing through it and the resistance (R) of the resistor. Mathematically, Ohm's law is expressed as:
V = IR
Given: I = 1A (current) R = 4Ω (resistance)
Using Ohm's law, we can calculate the potential difference (V):
V = (1A) * (4Ω) = 4V
Therefore, the potential difference across the resistor is 4V.
The power dissipated by a resistor can be calculated using the formula:
P = IV
Given: I = 1A (current) V = 4V (potential difference across the resistor)
Using the formula, we can calculate the power dissipated by the resistor (P):
P = (1A) * (4V) = 4W
Therefore, the power dissipated by the resistor is 4W.
If the internal resistance of the battery is increased to 4Ω, it will affect the circuit's total resistance (R_tot). The total resistance can be calculated by adding the resistance of the external resistor (4Ω) and the internal resistance (4Ω):
R_tot = R + r_internal = 4Ω + 4Ω = 8Ω
Using Ohm's law, we can calculate the new potential difference across the resistor (V_new):
V_new = I * R_tot = (1A) * (8Ω) = 8V
Therefore, if the internal resistance of the battery is increased to 4Ω, the new potential difference across the resistor will be 8V.