Question

please give solution in detail. ​

Answer 1

Solution :

Four cells each of resistance 1 / 4 ohm are connected in series

The equivalent resistance for resistors connected in Series combination is given by ,

$\implies \mathtt{R_s = R_ 1 + R_2 + R_3 + R_4}$

Here ,

• R 1 = R 2 = R 3 = R 4 = R = 1/4

$\implies \mathtt{R_s = 4R}$

$\implies \mathtt{R_s = 4 \times \dfrac{1}{4} }$

$\implies \mathtt{R_s = 1 \: ohm}$

This combination is in series with the external resistors of resistance 1 ohm

$\implies \mathtt{R_{net}= R_{ext} + R_s}$

$\implies \mathtt{R_{net}=1 + 1}$

$\implies \mathtt{R_{net}= 2 \: ohm}$

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All the 4 cells each having a EMF of 1.5 V are connected in series with each other but one cell is reversed thus having opposite polarity

$\implies\mathtt{E_{net} = 3E -E}$

$\implies\mathtt{E_{net} = 2E}$

$\implies\mathtt{E_{net} = 2 \times 1.5}$

$\implies\mathtt{E_{net} = 3V}$

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we need to find the net current flowing through the circuit ,

we know that ,

$\implies\mathtt{I = \dfrac{E_{net}}{R_{net}}}$

$\implies\mathtt{I = \dfrac{3}{2}}$

$\implies\mathtt{I = 1.5 A}$

The current flowing through the circuit is 1.5 A