When a 12V battery (i.e. a battery of EMF 12V) is connected across a lamp with a resistance of 6.8 Ω the PD across the lamp is 10.2V. Find (a) the current through the lamp, (b) the internal resistance of the battery.
(a)
"I=VR=10.2*6.8=69.4\\:\\rm A"(b)
"r=\\frac{{\\cal E}-V}{I}=\\frac{12-10.2}{69.3}=0.026\\:\\Omega"
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