A battery has an EMF of 12 V and internal resistance r - Scottish Highers Physics - Question 12 - 2023
Question 12
A battery has an EMF of 12 V and internal resistance r. The battery is connected in a circuit as shown.
(a) The reading on the ammeter is 0.38 A.
(i) Determine the... show full transcript
Worked Solution & Example Answer:A battery has an EMF of 12 V and internal resistance r - Scottish Highers Physics - Question 12 - 2023
Step 1
Determine the terminal potential difference (t.p.d.) of the battery.
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Answer
To find the terminal potential difference (V), we use the formula:
V=E−Iimesr
Where:
E = EMF of the battery = 12 V
I = current = 0.38 A
r = internal resistance (which we will calculate later)
First, we calculate the total resistance in the circuit. The total resistance (R) in the circuit is:
R=R1+R2=16Ω+16Ω=32Ω
Using Ohm's law:
I=R+rE⇒r=IE−R
So:
V=IimesR=0.38imes24=9.1V
Step 2
Calculate the internal resistance r of the battery.
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Answer
Using the earlier relationship,
12=0.38imes(24+r)
Expanding this, we have:
12=9.12+0.38r
Rearranging to isolate r gives:
0.38r=12−9.12=2.88
So:
r=0.382.88≈7.6Ω
Step 3
Calculate the power dissipated by the internal resistance of the battery.
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Answer
The power (P) dissipated by the internal resistance can be calculated using:
P=I2r
Substituting the values:
P=(0.38)2×7.6=1.1W
Step 4
State whether the power dissipated by the internal resistance of the battery is greater than, equal to, or less than the value determined in (a)(iii).
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Answer
The power dissipated will be less than the previous value determined, given that:
The total circuit resistance increases,
Current decreases,
Internal resistance stays the same.
Thus, if the internal resistance remains unchanged and the overall resistance in the circuit increases, the current through the internal resistance decreases, resulting in less power dissipation.
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