Question

2. Four resistors of 2 ohm each are connected first in series and then in parallel with a battery.Find the ratio of electric current In series and parallel connection.

Answer 1

Answer:

Ratio of series current to parallel

= 1 : 8

Explanation:

Total resistance Rt

For series, Rt = 2+2+2+2 = 4ohms

For parallel, 1/Rt = 1/2 + 1/2 + 1/2 + 1/2

1/Rt = 4/2, Rt = 2/4 ohms.

If we use a 1V battery, then,

I = V/Rt

I = 1/4 = 0.25 ampere for series arrangement.

I = 1/0.5 = 2 ohms.

Ratio of current of series to parallel = 0.25 : 2

= 1 : 8

Answer 2

Answer:

The ratio of electric current in series and in parallel connection is 0.0625

Explanation:

Here we have for the series resistors

∑R$_s$ = R₁ + R₂ + R₃ + R₄ = 8 Ω

The current is gotten from

V = IR

I$_s$ = V/R$_s$ = V/8 Ω

For the parallel connected resistors we have

$\frac{1}{\Sigma R_p} = \frac{1}{R_1} +\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}$

Which gives

$\frac{1}{\Sigma R_p} = \frac{1}{2} +\frac{1}{2}+\frac{1}{2}+\frac{1}{2} = \frac{4}{2} = 2$

Therefore,

$\Sigma R_p = \frac{1}{2}=0.5\hspace{0.09cm} \Omega$

From which,

$I_p=\frac{V}{R_p}$ = V/0.5 Ω

The ratio between the series and parallel connection electric current  is therefore,

$\frac{I_s}{I_p} = \frac{\frac{V}{8\hspace{0.09cm}\Omega} }{\frac{V}{0.5\hspace{0.09cm}\Omega} } = \frac{V}{8\hspace{0.09cm}\Omega} \times \frac{0.5\hspace{0.09cm}\Omega}{V} = \frac{0.5\hspace{0.09cm}\Omega}{8\hspace{0.09cm}\Omega} = 0.0625$.