Two resistors, one of 8 Ω and the other of 24 Ω are connected in parallel. This combination is connected in series with a 14 Ω resistor and a 12 V battery. The current in the 8 Ω resistor is:

Concept:

• Ohm's law: It states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperature, remain constant.
• Mathematically, current-voltage can be written as V = IR, where V = Voltage, I = Current, R = Resistance
• For equivalent resistance in parallel combinations
  • \(\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\)

• For equivalent resistance in series combinations

  • ​R eq  = R 1  + R 2  + . . . . + R n

• ​Current Division rule:
  • ​
  • The current flow through the resistor R 1 , \(I_1=\frac{IR_2}{R_1+R_2}\)
  • The current flows through the resistor R 2 , \(I_2=\frac{IR_1}{R_1+R_2}\)

Calculation:

Given, Resistance, R 1  = 8 Ω , R 2  = 24 Ω,Battery, V = 12 volt

The resistor is connected in parallel combination,

In parallel combination, \(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}\)

\(\frac{1}{R_p}=\frac{1}{8}+\frac{1}{24}\)

R_P  = 6 Ω 

Now, the resistance R P is connected in series with 14Ω,

R_eq  = 6 + 14 = 20 Ω 

Current in the circuit, \(I=\frac{V}{R}=\frac{12}{20}=0.6~A\)

The current flowing in the 8 Ω is calculated as,

\(I_1=\frac{IR_2}{R_1+R_2}\)

\(I_1=\frac{0.6\times 24}{8+24}=0.45A\)

Hence, the current in the 8 Ω resistor is 0.45 A.