A large number of AA batteries with emf E=1.5 V and resistors with resistance R form an infinite circuit . The internal resistance of the batteries is r. Thevenin's theorem tells us that this infinite circuit can be replaced by a single source with emf E_eq, connected in
series with an equivalent resistor R_eq. Determine the equivalent emf E_eq in volts if r/R=1/2.
To determine the equivalent emf E_eq in volts, we need to apply Thevenin's theorem to the infinite circuit formed by the batteries and resistors.
Thevenin's theorem states that any linear electrical network with voltage sources, current sources, and resistors can be replaced by a single voltage source in series with an equivalent resistor. The equivalent resistor is the resistance seen from the output terminals when all the voltage and current sources in the original network are replaced by their internal resistances.
In this case, we have an infinite circuit with multiple AA batteries with emf E = 1.5 V and resistors with resistance R. The internal resistance of the batteries is r, and we are given that r/R = 1/2.
To determine the equivalent emf E_eq, we need to find the voltage across the output terminals when the circuit is open, i.e., no current flows through it. Since the batteries and resistors form an infinite circuit, we can consider a single battery-resistor pair and analyze it.
Let's consider one battery and its corresponding resistor in the circuit. The voltage across the terminals of this battery can be given by Ohm's Law as:
V = E - Ir,
where V is the voltage across the terminals, E is the emf of the battery, I is the current through the resistor, and r is the internal resistance of the battery.
Since this is an infinite circuit, the current I will be the same for all battery-resistor pairs. Let's assume this common current is I_0.
The total voltage across the battery terminals will be:
V_total = V + V + V + ... (infinite times)
V_total = (E - I_0 r) + (E - I_0 r) + (E - I_0 r) + ...
Simplifying, we would get:
V_total = ∞E - ∞I_0 r
Since we have an infinite number of battery-resistor pairs, the total voltage across the terminals will be infinite.
However, if we remove one battery-resistor pair from the circuit, the voltage across the terminals will be:
V_removed = (E - I_0 r) + (E - I_0 r) + (E - I_0 r) + ...
Now, we have (N-1) pairs remaining, where N is a very large number. Since N is very large, we can approximate this sum as:
V_removed ≈ (N-1)(E - I_0 r)
In the above equation, I_0 is the current through the resistor. We need to determine the value of I_0 such that removing one pair from the circuit would give us an equivalent circuit with the same voltage across the terminals.
In an infinite circuit with constant current, the voltage across each resistor is given by Ohm's Law:
V = IR,
where I is the current and R is the resistance.
Since r/R = 1/2, the internal resistance of the battery, r, is half the resistance of each resistor in the circuit.
We can rewrite Ohm's Law for the resistor as:
E - I_0 r = I_0 (2R)
Simplifying, we find:
I_0 = E / (3R)
Substituting this value of I_0 in the equation for V_removed, we have:
V_removed ≈ (N-1)(E - E / (3R) * r)
Simplifying further, we get:
V_removed ≈ (N-1)(E - E / 3)
Let's assume that V_removed is equal to the desired equivalent emf E_eq, which is the voltage across the output terminals when the circuit is open:
E_eq ≈ (N-1)(E - E / 3)
Since N is a very large value, we can approximate (N-1) as N:
E_eq ≈ N(E - E / 3)
Simplifying, we find:
E_eq ≈ 2E / 3
Therefore, the equivalent emf E_eq is approximately 2/3 times the original emf (1.5V), which is about 1V.
Hence, the equivalent emf E_eq is approximately 1V.