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?
I = V/Rm = 5.70/95 = 0.06A.
r = V/I = (6-5.7)/0.06 = 0.30/0.06 = 5. Ohms.
To determine the internal resistance of the battery, we can use the concept of potential difference and an understanding of how the voltmeters work.
First, let's establish some basic facts relevant to this problem:
1. The voltage across the terminals of a battery corresponds to its electromotive force (EMF), which is the maximum potential difference it can produce without any external load connected.
2. An analogue voltmeter is connected in parallel to the battery, and it has a resistance of 95 ohms.
3. A digital meter has a very high resistance, which is typically in the order of megaohms (10^6 ohms).
Now, let's proceed with the calculation:
The potential difference measured by the analogue voltmeter is given as 5.70 V, and its internal resistance is 95 ohms. We'll denote the EMF of the battery as E.
According to Ohm's Law, the voltage measured by the voltmeter is given by:
V_analogue = E - I_analogue * R_internal
Here, I_analogue represents the current flowing through the voltmeter, and R_internal is the internal resistance of the battery.
Similarly, the potential difference measured by the digital meter is given as 6.00 V, and the internal resistance of the digital meter is extremely high. We can neglect the current flowing through the digital meter due to its high resistance.
Thus, we can write:
V_digital = E
Now we have a system of two equations:
V_analogue = E - I_analogue * R_internal (Equation 1)
V_digital = E (Equation 2)
Substituting the given values, we get:
5.70 = E - I_analogue * 95 (Equation 3)
6.00 = E
From Equation 2, we find that E = 6.00 V. Substituting this value into Equation 3, we get:
5.70 = 6.00 - I_analogue * 95
Rearranging the equation to isolate I_analogue, we have:
I_analogue = (6.00 - 5.70) / 95
Simplifying this expression, we find:
I_analogue = 0.03 A
Now that we have the current flowing through the voltmeter, we can substitute it back into Equation 1 to determine the internal resistance:
5.70 = 6.00 - (0.03 * R_internal)
Rearranging the equation, we find:
0.30 = 0.03 * R_internal
Dividing both sides by 0.03, we get:
R_internal = 0.30 / 0.03
Simplifying this expression, we find:
R_internal = 10 ohms
Therefore, the internal resistance of the battery is 10 ohms.