Three resistors of 4 Ω, 8 Ω and 6 Ω are connected in parallel combination with a 9 V battery. The total current drawn, from the battery is: 

The correct answer is 4.8 A.

Key Points

Concept:

• Resistors in Series:

                     Two or more resistors are said to be connected in series when the same amount of current flows through all the resistors. In such circuits, the voltage across each resistor is different.

                   Rtotal = R1 + R2 + ….. + Rn

• Resistors in Parallel:

               Two or more resistors are said to be connected in parallel when the voltage is the same across all the resistors, but different amounts of current flow throw all the resistors.

                \(\frac{1}{R_{total}}= \frac{1}{R_{1}} + \frac{1}{R_{2}}+ \dotsb + \frac{1}{R_{n}}\)

• Total Current in a circuit:

                 According to Ohm's law, the current is the ratio of the potential difference and the resistance. Thus, the current formula is given :

                 \(I = \frac{V}{R_{eq}}\)

Calculation:

As in the given question, the resistors are connected in parallel, thus the equivalent resistance is :

      \(\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{8}+\frac{1}{6}\)

      \(\frac{1}{R_{eq}} = \frac{1}{0.545}\)

     \(R_{eq} = 1.84\)  ohm

Now, We have V = 9V and R = 1.84,

Thus, Total Current in the circuit is :

     \(I = \frac{V}{R} \)

     \(I = \frac{9}{1.84}\)

   \(I= 4.8 A\)

Hence, The correct answer is: '4.8A'