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ICSE Class 10 Physics 2016 Paper

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A battery of emf 12 V12\ \mathrm{V} and internal resistance 2 Ω2\ \Omega is connected with two resistors AA and BB of resistance 4 Ω4\ \Omega and 6 Ω6\ \Omega respectively joined in series.
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Question : 63 of 73
Marks: +1, -0
The potential difference across 6Ω6 \Omega Resistor.
Solution:  
Potential difference across 6Ω=IR6 \Omega = IR
=1×6=6V=1 \times 6 = 6\,\text{V}
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