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ICSE Class 10 Physics 2015 Paper

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A cell of Emf 2 V2\ \mathrm{V} and internal resistance 1.2 Ω1.2\ \Omega is connected with an ammeter of resistance 0.8 Ω0.8\ \Omega and two resistors of 4.5 Ω4.5\ \Omega and 9 Ω9\ \Omega as shown in the diagram below :
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Question : 60 of 74
Marks: +1, -0
What is the potential difference across the terminals of the cell?
Solution:  
     Terminal Voltage (V)   \;\;\text{ Terminal Voltage (V) }\;
=EIr=20.4×1.2= E - I r = 2 - 0.4 \times 1.2
  =20.48\;=2-0.48
  =1.52   volt. \;=1.52 \;\text{ volt. }
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