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ICSE Class 10 Physics 2022 Paper

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In the above circuit diagram, calculate :
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Question : 16 of 45
Marks: +1, -0
the external resistance of the circuit
Solution:  
As, 12Ω12 \Omega and 8Ω8 \Omega are in parallel, so
    1RP=  112+  18\;\;\frac{1}{R_P}=\;\frac{1}{12}+\;\frac{1}{8}
⇒      1RP=  2+324=  524\Rightarrow \;\;\;\frac{1}{R_P}=\;\frac{2+3}{24}=\;\frac{5}{24}
⇒    RP=  245=4.8Ω\Rightarrow \;\; R_P=\;\frac{24}{5}=4.8 \Omega
Now, RPR_P and 2Ω2 \Omega are in series, so
Req=4.8+2=6.8ΩR_{eq}=4.8+2=6.8 \Omega
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