A high-resistance voltmeter reads 3.0V when connected across the terminals of a battery on open circuit and 2.6V when the battery supplies a current of 0.2A through a lamp. The resistance of lamp is
2.00\(\Omega\)
13.00\(\Omega\)
0.52\(\Omega\)
0.13\(\Omega\)
1.50\(\Omega\)
Explanation
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Discussions (12)
The voltage which is V, from ohms law V= IR.
From I=E/R+r. By cross multiplying, E= IR + Ir, where IR=V and Ir is referred to as lost voltage.
From the question above, the voltmeter reads 3.0v when current hasn't passed through it but reads 2.6v when a current of 0.2A passes through it, which means there was lost in voltage, so the lost in voltage is given as
Ir= 3.0-2.6= 0.4v
The question goes thus: THE RESISTANCE OF THE LAMP, automatically, the lost voltage is not in used which means 2.6v is active and it's demoted as R
From ohm's law, V=IR
V=2.6, I= 0.2, R=?
I=V/R
I=2.6/0.2= 13Ω
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connected... open circuit, i.e in emf
emf = 3.0
V = 2.6
I = 0.2...
Resistance = ?
and resistance is V / I
R = v / I
R = 2.6 / 0.2
R = 13...
just to confuse you, jamb put emf and that is just to find lot voltage then you ust the lost voltage to get your internal resistance as at;
emf = V + v
3 = 2.6 + v
v = 3 - 2. 6
v = 0.4 therefore
internal resistance = lost voltage / current
r = v / I
r = 0.4 / 0.2
r = 2..... as you can see is in the option, so therefore its wrong.... RESISTANCE not •internal resistance•
