Question

The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -

Answer 1

As discussed in @8173

Req = \frac{40\times 120}{40+120}\:\Omega

I = \frac{V}{R} =\frac{7}{5+\frac{40\times 120}{40+120}}

I = \frac{7}{5+30} \Rightarrow I = \frac{1}{5} = 0.2A

Answer 2

The current drawn from the cell will be 0.2 A.

Explanation:

The P and Q arms of Wheatstone`s bride are in series,

so $R_1=10+30$

          = 40 ohm

Similarly, R and S are also in series,

so   $R_2=30+90$

And both are in parallel so equivalent resistance,

$R_e_q=\frac{1}{R_1}+\frac{1}{R_2}$

$R_e_q=\frac{40\times120}{120+40} =$

      = 30 ohm.

As internal resistance of the cell is 5 ohm.

Therefore net resistance,

$R_n_e_t=30+5$

        = 35 ohm.

The current drawn from the cell,

$i=\frac{7}{35}$

i  = 0.2 A.

Thus, current drawn from the cell will be 0.2 A.

#Learn More:  combination of resistances

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