To find the voltage $V_{ac}$ across points $a$ and $c$, we need to calculate the total voltage drop across all the resistors in the circuit.

1. Analyze the Circuit: The circuit is a series circuit with resistors $R_1 = 470 \, \Omega$, $R_2 = 180 \, \Omega$, and $R_3 = 390 \, \Omega$ in series with a $24 \, V$ voltage source.

2. Total Resistance $R_{total}$: Since the resistors are in series, the total resistance is the sum of the resistances: $R_{total} = R_1 + R_2 + R_3 = 470 \, \Omega + 180 \, \Omega + 390 \, \Omega = 1040 \, \Omega$

3. Calculate the Current $I$: Using Ohm's law $V = IR$, the current $I$ through the circuit is: $I = \frac{V}{R_{total}} = \frac{24 \, V}{1040 \, \Omega} \approx 0.0231 \, A$

4. Voltage Across $R_1$, $R_2$, and $R_3$:

   • Voltage across $R_1$: $V_{R1} = I \cdot R_1 = 0.0231 \, A \cdot 470 \, \Omega \approx 10.857 \, V$
   • Voltage across $R_2$: $V_{R2} = I \cdot R_2 = 0.0231 \, A \cdot 180 \, \Omega \approx 4.158 \, V$
   • Voltage across $R_3$: $V_{R3} = I \cdot R_3 = 0.0231 \, A \cdot 390 \, \Omega \approx 9.009 \, V$

5. Voltage $V_{ac}$: Since the total voltage is the sum of the voltage drops across $R_1$, $R_2$, and $R_3$, we find that: $V_{ac} = V_{R1} + V_{R2} + V_{R3} = 10.857 \, V + 4.158 \, V + 9.009 \, V = 24 \, V$

Therefore, the voltage $V_{ac}$ is indeed 24 V, as it is the total voltage supplied by the source.

Would you like more details or have any questions?

Related Questions:

1. What would happen to $V_{ac}$ if $R_2$ were doubled?
2. How would the total current change if $R_3$ were removed?
3. Can we calculate the power dissipated by each resistor?
4. What effect would increasing the supply voltage have on $V_{ac}$?
5. How would the current change if an additional resistor were added in series?

Tip:

In a series circuit, the current is constant across all elements, and the total voltage is the sum of individual voltage drops across each resistor.