Two identical light bulb, each with resistance R = 2 are connected to a source with E = 8
V and negligible internal resistance. Find the current through each bulb. The potential
difference across each bulb and the power delivered to each bulb and to the entire network
if the bulbs are connected parallel
Given:
"E=8V; R=2\\Omega;"
Find:
"I1" - current though 1st bulb;
"I2" - current though 2nd bulb;
"U1" - potential difference across 1st bulb;
"U2" - potential difference across 2nd bulb;
"P1" - power delivered to 1st bulb;
"P2" - power delivered to 2nd bulb;
"P" - total power delivered to system.
Solution:
Let's find total resistance and current.
The bulbs connected parallel, so
"Rt=R\/2=2\\Omega\/2=1\\Omega;" "I=E\/Rt=8V\/1\\Omega=8A;"
Bulbs have the same resistance and connected parallel, so from Krichhoff's rule:
"I1+I2=I;"
"I1=I2;" "I1=I2=I\/2=8A\/2=4A;"
Also,
"U1=U2=RI1=2\\Omega*4A=8V;"
"P1=P2=I1U1=4A*8V=32W;"
"P=EI=8V*8A=64W."
Answer:
"I1=I2=4A;"
"U1=U2=8V;"
"P1=P2=32W;"
"P=64W."
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