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CBSE Class 10 Science 2015 Term I Set 1

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Question : 16 of 36
Marks: +1, -0
Find the current drawn from the battery by the network of four resistors shown in the figure.
Solution:  
Resultant resistance of R1,R2R_1, R_2 and R3R_3 :
R′=R1+R2+R3R^{'} = R_1+R_2+R_3
=10+10+10=30Ω= 10+10+10 = 30 \Omega
Resultant resistance of the circuit :
1R=1R′+1R4\frac{1}{R} = \frac{1}{R^{'}} + \frac{1}{R_4}
=130+110=1+330=430= \frac{1}{30} + \frac{1}{10} = \frac{1+3}{30} = \frac{4}{30}
R=304=152=7.5R = \frac{30}{4} = \frac{15}{2} = 7.5
Here, V=3 V,I=V = 3 \text{ V}, I = ?
By Ohm's law,
⇒V=IR\Rightarrow V = I R
I=VR=37.5=3075=0.4 AI = \frac{V}{R} = \frac{3}{7.5} = \frac{30}{75} = 0.4 \text{ A}
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