A cell of e.m.f. 1.5V is connected in series with a resistor of resistance 39. A high-resistance voltmeter connected across the cell registers only 0.9V. Calculate the internal resistance of the cell
1.85Ω
2.0Ω
2.4Ω
4.5Ω
5.0Ω
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Using E=V+v=I(R+r)
(Check SlimmieRex's comment 2 no wot diz symbols represent)
E=1.5v,R=3A,V=0.9v
since,E=V+v, v=E-V=1.5-0.9=0.6v
I=V/R=0.9/3=0.3A
Sub d known into d eqn E=I(R+r)
1.5=0.3(3+r)
which gives r=2ohms
i dont understand buh d formular of internal resistance is V = E - Ir
Where:
V = pd across the external circuit (V)
E = emf of the cell (V)
I = current through the cell (A)
r = value of the internal resistance (Ω)
( Ir = the p.d. across the internal resistor)
i’ve been trying to solve it and i am getting 26 ohms
i didn’t know they meant 3 
i’ve been trying to solve it for an hour
Should the emf not be 3, because it is connected in series? Why is everyone using 1.5v?
