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CHECK YOURSELF ANSWERS
1The charge transferred is 1 C which corresponds to 11.6Ã10â19=6.2Ã1018 electrons.
2R=ÏLA. L is doubled; the radius is doubled so the area increases by 4. Hence the new resistance is Râ²=Ï2L4A=R2.
3P=V2R
So, 32=122RâR=12232
The new power is:
Pâ²=V2R=152(12232)=152122Ã32=5242Ã32=50Â W.
4The total resistance increases so the current decreases.
5We are told that 1R+1R+1R=11 which means that R = 3 Î©. So the series connection would give a total resistance of 9 Î©.
6The total resistance in the series connection is 2R. In the parallel connection it is R2. Since resistance is reduced by 4 the power will go up by 4 (sinceÂ P=V2R) and so the new power will be 64 W.
7The total resistance of the circuit is 6.0 Î© and so the current is I=6.03.0=2.0Â A.
aThe power dissipated in the cell is P = rI2 = 0.50 Ã 2.02 = 2.0 W.
bThe power in the external resistor is P = RI2 = 2.5 Ã 2.02 = 10 W. (The total power is P = ÎµI = 6.0 Ã 2.0 = 12 W (= 10 + 2.0) as expected.)
