i=2.5500=5mA
VCD=5×10−3×550
=2.75
Voltage divide rule –
VCD=12×(550||R)550||R+550
2.75=12×550×R550×R+550(R+550) (2.75)(2R+550)=12R
Two identical cells, each of emf 2 V and of unknown internal resistance r are connected in parallel. This combination is connected to a 4 ohm resistor. If the terminal voltage across the cells in the closed circuit is 1.5 V then the internal resistance of the cell is
A cell of internal resistance 1Ω is connected across a resistor. A voltmeter having variable resistance G is used to measure p.d across resistor. The plot of voltmeters reading V against G is shown. What is value of external resistor R (in ohm)?
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