A certain battery has a 12.0-V emf and an internal resistance of 0.100Ω0.100Ω size 12{0 "." "100" %OMEGA } {}. (a) Calculate its terminal voltage when connected to a 10.0-Ω10.0-Ω size 12{"10" "." 0- %OMEGA } {} load. (b) What is the terminal voltage when connected to a 0.500-Ω0.500-Ω size 12{0 "." "500-" %OMEGA } {} load? (c) What power does the 0.500-Ω0.500-Ω size 12{0 "." "500-" %OMEGA } {} load dissipate? (d) If the internal resistance grows to 0.500 Ω0.500 Ω size 12{0 "." "500 " %OMEGA } {}, find the current, terminal voltage, and power dissipated by a 0.500-Ω0.500-Ω size 12{0 "." "500-" %OMEGA } {} load.
Strategy
The analysis above gave an expression for current when internal resistance is taken into account. Once the current is found, the terminal voltage can be calculated using the equation V=emf−IrV=emf−Ir size 12{V="emf" - ital "Ir"} {}. Once current is found, the power dissipated by a resistor can also be found.
Solution for (a)
Entering the given values for the emf, load resistance, and internal resistance into the expression above yields
I=emfRload+r=12.0 V10.1Ω=1.188 A.I=emfRload+r=12.0 V10.1Ω=1.188 A. size 12{I= { {"emf"} over {R rSub { size 8{"load"} } +r} } = { {"12" "." 0" V"} over {"10" "." "1 " %OMEGA } } =1 "." "188"" A"} {}
(3)Enter the known values into the equation V=emf−IrV=emf−Ir size 12{V="emf" - ital "Ir"} {} to get the terminal voltage:
V
=
emf−
Ir
=
12.0 V
−
(1.188 A)
(0.100 Ω)
=
11.9 V.
V
=
emf−
Ir
=
12.0 V
−
(1.188 A)
(0.100 Ω)
=
11.9 V.
(4)Discussion for (a)
The terminal voltage here is only slightly lower than the emf, implying that 10.0Ω10.0Ω size 12{"10" "." 0 %OMEGA } {} is a light load for this particular battery.
Solution for (b)
Similarly, with Rload=0.500ΩRload=0.500Ω size 12{R rSub { size 8{"load"} } =0 "." "500"` %OMEGA } {}, the current is
I=emfRload+r=12.0 V0.600Ω=20.0 A.I=emfRload+r=12.0 V0.600Ω=20.0 A. size 12{I= { {"emf"} over {R rSub { size 8{"load"} } +r} } = { {"12" "." 0" V"} over {0 "." "600 " %OMEGA } } ="20" "." 0" A"} {}
(5)The terminal voltage is now
V
=
emf−Ir=12.0 V−(20.0 A)(0.100 Ω)
=
10.0 V.
V
=
emf−Ir=12.0 V−(20.0 A)(0.100 Ω)
=
10.0 V.
(6)Discussion for (b)
This terminal voltage exhibits a more significant reduction compared with emf, implying 0.500 Ω0.500 Ω size 12{0 "." "500 " %OMEGA } {} is a heavy load for this battery.
Solution for (c)
The power dissipated by the 0.500 - Ω0.500 - Ω size 12{0 "." "500-" %OMEGA } {} load can be found using the formula P=I2RP=I2R size 12{P=I rSup { size 8{2} } R} {}. Entering the known values gives
Pload=I2Rload=
(20.0 A
)
2
(0.500 Ω)=2.00×102W.Pload=I2Rload=
(20.0 A
)
2
(0.500 Ω)=2.00×102W. size 12{P rSub { size 8{"load"} } =I rSup { size 8{2} } R rSub { size 8{"load"} } = \( "400"" A" rSup { size 8{2} } \) \( 0 "." "500" %OMEGA \) ="200"" W"} {}
(7)Discussion for (c)
Note that this power can also be obtained using the expressions V2RV2R size 12{ { {V rSup { size 8{2} } } over {R} } } {} or IVIV size 12{ ital "IV"} {}, where VV size 12{V} {} is the terminal voltage (10.0 V in this case).
Solution for (d)
Here the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load resistance. As before, we first find the current by entering the known values into the expression, yielding
I=emfRload+r=12.0 V1.00 Ω=12.0 A.I=emfRload+r=12.0 V1.00 Ω=12.0 A. size 12{I= { {"emf"} over {R rSub { size 8{"load"} } +r} } = { {"12" "." 0" V"} over {1 "." "00 " %OMEGA } } ="12" "." 0" A"} {}
(8)Now the terminal voltage is
V
=
emf−Ir=12.0 V−(12.0 A)(0.500 Ω)
=
6.00 V,
V
=
emf−Ir=12.0 V−(12.0 A)(0.500 Ω)
=
6.00 V,
(9)and the power dissipated by the load is
Pload=
I2
Rload
=
(12.0 A)2
(0.500Ω)=72.0 W.Pload=
I2
Rload
=
(12.0 A)2
(0.500Ω)=72.0 W. size 12{P rSub { size 8{"load"} } =I rSup { size 8{2} } R rSub { size 8{"load"} } = \( "144"" A" rSup { size 8{2} } \) \( 0 "." "500" %OMEGA \) ="72" "." 0" W"} {}
(10)Discussion for (d)
We see that the increased internal resistance has significantly decreased terminal voltage, current, and power delivered to a load.