When potential difference across the terminals of a battery is measured using an analogue voltmeter of resistance 95 ohms, the reading on the voltmeter is 5.70 V. When it is measured using a very high resistance digital meter the reading is 6.00 V. What is the internal resistance of the battery?
in the 1st case ... the internal voltage drop is 0.30 v
... the current is ... 5.70 v / 95 Ω
internal resistance ... 0.30 v / current
Oh I think I understand, when I solve it I get 5 ohms as an answer, is that correct?
Thank you but I’m still a little confused.. what would the current be in this case?
To find the internal resistance of the battery, we can use the formula:
Internal Resistance (r) = (V1 - V2) * R2 / (V2 - V3)
Where:
- V1 is the reading on the analogue voltmeter (5.70 V)
- V2 is the reading on the digital meter (6.00 V)
- V3 is the reading on the voltmeter if the internal resistance of the battery was zero (this is unknown)
- R2 is the resistance of the digital meter (very high resistance)
First, we need to calculate V3. We can do this using Ohm's law:
V3 = V2 * (R1 + R2) / (R2)
Where:
- R1 is the resistance of the analogue voltmeter (95 ohms)
- R2 is the resistance of the digital meter (very high resistance)
Substituting the given values:
V3 = 6.00 V * (95 ohms + very high resistance) / (very high resistance)
Since the resistance of the digital meter is very high, we can consider it as approaching infinity. Therefore, the equation simplifies to:
V3 = 6.00 V * (95 ohms + infinity) / infinity
This results in V3 being approximately equal to 6.00 V.
Now, we can substitute the values into the formula to calculate the internal resistance:
r = (V1 - V2) * R2 / (V2 - V3)
= (5.70 V - 6.00 V) * (very high resistance) / (6.00 V - 6.00 V)
= -0.30 V * (very high resistance) / 0 V
Since any value multiplied by 0 is equal to 0, the numerator becomes 0. Therefore, the internal resistance of the battery is 0 ohms.