Question

Two 40 ohm resistor and a 20 ohm resistor are connected in a parallel with a 12V battery. Calculate their effective resistance and the current through each resistor and what is the I flowing through the battery?

Answer 1

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Answer 2

Answer:

The effective resistance and the current through each resistor and the I flowing through the battery are $\frac{40}{3}\Omega$,$\frac{3}{10} A$, $\frac{3}{5} A$ and $\frac{9}{10} A$ respectively.

Explanation:

The effective resistance in the circuit is given as,

$\frac{1}{R} =\frac{1}{R_1}+ \frac{1}{R_2}$      (1)

From the question we have,

R₁=40

R₂=20

The voltage of the battery=12V

By substituting the value of R₁ and R₂ in equation (1) we get;

$\frac{1}{R} =\frac{1}{40}+ \frac{1}{20}$

$\frac{1}{R} =\frac{1+2}{40}$

$R=\frac{40}{3}\Omega$        (2)

The current(I) flowing through the battery is given as,

$I=\frac{V}{R}$

$I=\frac{12}{\frac{40}{3} }$

$I=\frac{12\times 3}{40}$

$I=\frac{9}{10} A$     (3)

Since the resistors, R₁ and R₂ are connected in parallel the voltage across them will remain the same.i.e.12V.So, the current through them will be given as,

$I_1=\frac{V}{R_1}$

$I_1=\frac{12}{40} =\frac{3}{10} A$     (4)

$I_2=\frac{V}{R_2}$

$I_2=\frac{12}{20}=\frac{3}{5} A$        (5)

Hence, effective resistance and the current through each resistor and the I flowing through the battery are $\frac{40}{3}\Omega$,$\frac{3}{10} A$, $\frac{3}{5} A$ and $\frac{9}{10} A$ respectively.