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ICSE Class 10 Physics 2023 Paper

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Find the value of current I drawn from the cell.
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Question : 68 of 76
Marks: +1, -0
Calculate the current I.
Solution:  
In the given circuit diagram, the two 15 Ω15\ \Omega resistor are in series, so their total resistance
Rs=15+15=30 ΩR_{\mathrm{s}} = 15+15 = 30\ \Omega
Now, RsR_s and the given 30 Ω30\ \Omega resistors are in parallel, so the net resistance
  1Rp  =  130+  130\; \frac{1}{R_p} \; = \; \frac{1}{30} + \; \frac{1}{30}
  1Rp  =  1+130=  230\; \frac{1}{R_p} \; = \; \frac{1+1}{30} = \; \frac{2}{30}
∴    Rp  =  302=15 Ω\therefore \; \; R_p \; = \; \frac{30}{2} = 15\ \Omega
So,     Req=15 Ω\; \; R_{\mathrm{eq}} = 15\ \Omega
We know,     I=εReq+r\; \; I = \frac{\varepsilon}{R_{\mathrm{eq}} + r}
∴    I=3.415+2=3.417=0.2 A\therefore \; \; I = \frac{3.4}{15+2} = \frac{3.4}{17} = 0.2\ \mathrm{A}
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