Use the superposition theorem to find the current through R_{2} of Figure 8-17.
Use the superposition theorem to find the current through R_{2} of Figure 8-17.
Step 1: Replace V_{S2} with a short and find the current through R_{2} due to voltage source V_{S1} , as shown in Figure 8-18. To find I_{2}, use the current-divider formula (Equation 6-6 : I_{\chi }= \left(\frac{R_{T}}{R_{\chi}} \right) I_{T}). Looking from V_{S1},
R_{T(S1)}= R_{1}+ \frac{R_{3}}{2} = 100 \ \Omega + 50 \ \Omega = 150 \ \OmegaI_{T(S1)}= \frac{V_{S1}}{R_{T(S1)}} = \frac{10 \ V}{150 \ \Omega } = 66.7 \ mA
The current through R_{2} due to V_{S1} is
I_{2(S1)}= \left(\frac{R_{3}}{R_{2}+ R_{3}} \right) I_{T(S1)} = \left(\frac{100 \ \Omega }{200 \ \Omega } \right) 66.7 \ mA = 33.3 \ mANote that this current is upward through R_{2}.
Step 2: Find the current through R_{2} due to voltage source V_{S2} by replacing V_{S1} with a short, as shown in Figure 8-19. Looking from V_{S2}.
R_{T(S2)}= R_{3}+ \frac{R_{1}}{2} = 100 \ \Omega + 50 \ \Omega = 150 \ \OmegaI_{T(S2)}= \frac{V_{S2}}{R_{T(S2)}} = \frac{5 \ V}{150 \ \Omega } = 33.3 \ mA
The current through R_{2} due to V_{S2}is
I_{2(S2)}= \left(\frac{R_{1}}{R_{1}+ R_{1}} \right) I_{T(S2)} = \left(\frac{100 \ \Omega }{200 \ \Omega } \right) 33.3 \ mA = 16.7 \ mANote that this current is upward through R_{2}.
Step 3: Both component currents are upward through R_{2}, so they have the same algebraic sign. Therefore, add the values to get the total current through R_{2}.
I_{2(tot)}=I_{2(s1)} +I_{2(s2)} = 33.3 mA + 16.7 mA = 50 mA