Question
Figure 21.70 In this circuit, assume the currents through $R_1$, $R_2$ and $R_3$ are $I_1$, $I_2$ and $I_3$ respectively and all are flowing in the clockwise direction.
1. Find the equation obtained by applying Kirchhoff’s junction rule at point A.
2. Find the equations obtained by applying Kirchhoff’s loop rule in the upper and lower loops.
3. Assume $R_1 = R_2 = 6 \Omega$, $R_3 = 12 \Omega$, $r_1 = r_2 = 0 \Omega$, $\xi_1 = 6 \textrm{ V}$ and $\xi_2 = 4 \textrm{ V}$. Calculate $I_1$, $I_2$ and $I_3$.
4. For the situation in which $\xi_2$ is replaced by a closed switch, repeat parts (a) and (b). Using the values for $R_1$, $R_2$, $R_3$, $r_1$ and $\xi_1$ from part (c) calculate the currents through the three resistors.
5. For the circuit in part (d) calculate the output power of the voltage source and across all the resistors. Examine if energy is conserved in the circuit.
6. A student implemented the circuit of part (d) in the lab and measured the current though one of the resistors as 0.19 A. According to the results calculated in part (d) identify the resistor(s). Justify any difference in measured and calculated value.
Question Image
1. $I_1 + I_3 = I_2$
2. see video
3. $I_1 = \dfrac{8}{15}$, $I_2 = \dfrac{7}{15}$, $I_3 = \dfrac{-1}{15}$
4. $I_1 = \dfrac{2}{5}$, $I_2 = \dfrac{3}{5}$, $I_3 = \dfrac{1}{5}$
5. $P_{total} = \dfrac{18}{5} \textrm{ W}$, $P_{R_1} = \dfrac{24}{25} \textrm{ W}$, $P_{R_2} = \dfrac{54}{25} \textrm{ W}$, $P_{R_3} = \dfrac{12}{25} \textrm{ W}$
Yes, energy is conserved.
6. The resistor is $R_3$.
The ammeter introduces additional resistance, $r$, in series with $R_3$, thereby reducing the current.
Solution Video