Question

Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. This combination is connected to a battery of emf 20 V and negligible internal resistance. The total current drawn from the battery is?

Answer 1

Hint: When many resistors are connected in parallel, the voltage across each resistor is the same. But the current across every resistor is different when the circuit is connected to external emf. The maximum current opts for a path of least resistance.

Complete step by step answer:
Refer to the circuit given below.

Let the current through $2\Omega, 3\Omega, 5\Omega$ resistors are ${I_1},{I_2},{I_3}$ respectively.
Current ${I_1}$ flowing through a resistor ${R_1}$ is given by:
${I_1} = \dfrac{E}{{{R_1}}} \\\ {I_1} = \dfrac{{20\;{\text{V}}}}{{2\;\Omega }} \\\ {I_1} = 10\;{\text{A}} \\\$
Current ${I_2}$ flowing through a resistor ${R_2}$ is given by:
${I_2} = \dfrac{E}{{{R_2}}} \\\ {I_2} = \dfrac{{20\;{\text{V}}}}{{4\;\Omega }} \\\ {I_2} = 5\;{\text{A}} \\\$
Current ${I_3}$ flowing through a resistor ${R_3}$ is given by:
${I_3} = \dfrac{E}{{{R_3}}} \\\ {I_3} = \dfrac{{20\;{\text{V}}}}{{5\;\Omega }} \\\ {I_3} = 4\;{\text{A}} \\\$
Therefore, the total current drawn from the battery is given by:
${I_t} = {I_1} + {I_2} + {I_3} \\\ {I_t} = 10\,{\text{A}} + 5\,{\text{A}} + 4\,{\text{A}} \\\ {I_t} = 19\,{\text{A}} \\\$
Hence the total current flowing is 19 A.

Additional information: When a current is flowing in a circuit then there is a voltage drop in source voltage or battery and it is termed as internal resistance. The electrolytic material inside the battery is responsible for internal resistance.

Note: When the internal resistance is given, the current through a resistor is given by $I = \dfrac{E}{{R + r}}$ where r is the internal resistance offered by the battery itself. It can be seen that the current through a resistor is reduced when an internal resistance is there.