Answer:
a) 0.73A
b) 0.23A
c) 2.76V
Explanation:
We need to first resolve the two resistors in series. The resistors in series are
12 Ω and 3 Ω AND 6 Ω and 3 Ω
For 12 Ω and 3 Ω in series, total effective resistance = 12 Ω + 3 Ω = 15 Ω
For 6 Ω and 3 Ω in series, total effective resistance = 6 Ω + 3 Ω = 9 Ω
Since the equivalent series resistors i.e 15Ω ND 9Ω are connected in parallel, the total effective resistance Rt will be expressed as;
1/Rt = 1/15+1/9
1/Rt = (3+5)/45
1/Rt = 8/45
Rt = 45/8 Ω
a) If a potential difference of 4 V is applied across it, the total current I in the circuit can be derives using the ohms law.
According to the law E = IRt
Given E = 4V, Rt = 45/8
I = E/Rt
I = 4/(45/8)
I = 4 * 8/45
I = 32/45
I = 0.73A
Hence, the current drawn from the battery is 0.73A
b) Before we can calculate the current in the 12Ω resistor, we need to calculate the current in the equivalent resistance of 15Ω(sum of 12Ω and 3Ω)
Current in the 15Ω resistor = Voltage across the 15Ω resistor/Resistance
Current in the 15Ω resistor = 4/15
Current in the 15Ω resistor = 0.27A
Since the same current flows in a series connected resistors, hence the correct in the 12ohms resistor is also 0.27A.
c) Before we can calculate the pd across the 6ohms resistor, we need to know the voltage across the effective resistance of 9ohms(6ohms+3ohms). The pd across the 9ohm resistance will be the same as the source voltage i.e 4Volts.
We will need to share this 4volts between the 6ohms and the 3ohms using ohms law.
According to the law, V = IR
For the 6ohms resistor, voltage across it will be;
V = (0.73-0.27)×6
V= 0.46×6
V = 2.76Volts.
Hence the voltage across the 6ohms resistor is 2.76V.
