$\text { Figure } 6 \text { '1 }$ -If R₁=4.7$\mathrm { k } \Omega$, R₂=3.3$\mathrm { k } \Omega$, R₃=1 $\mathrm { k } \Omega$ and V_S=50 V in Figure 6-1,I_T equals_____ A) 75.7 mA B) 9.15 mA C) 12.1 mA D) 8.8 mA

Question

Quizplus · Accepted Answer

Using voltage divider formulas, we can calculate the voltage at node S1U1B1:

$$V_{S1U1B1}=V_{S1U1B1S1U1B0}\cdot\frac{R_{S1U1B12S1U1B0}}{R_{S1U1B11S1U1B0}+R_{S1U1B12S1U1B0}}=50	ext{ V}\cdot\frac{3.3	ext{ k}\Omega}{4.7	ext{ k}\Omega+3.3	ext{ k}\Omega}=22.64	ext{ V}$$

Then, we can use Ohm's law to find IS1U1B1TS1U1B0:

$$I_{S1U1B1T,S1U1B0}=\frac{V_{S1U1B1}}{R_{S1U1B13S1U1B0}}=\frac{22.64	ext{ V}}{1	ext{ k}\Omega}=22.64	ext{ mA}$$

However, we need to convert this to IS1U1B1TS1U1B0, which can be done using KCL:

$$I_{S1U1B1T,S1U1B0}=I_{S1U1B1T,S1U1B2}+I_{S1U1B1T,S1U1B1}$$

We already know that $I_{S1U1B1T,S1U1B2}=3.3	ext{ mA}$ (from the voltage divider equation for RS1U1B12S1U1B2), so we can solve for $I_{S1U1B1T,S1U1B1}$:

$$I_{S1U1B1T,S1U1B1}=I_{S1U1B1T,S1U1B0}-I_{S1U1B1T,S1U1B2}=22.64	ext{ mA}-3.3	ext{ mA}=19.34	ext{ mA}$$

Therefore, the answer is B, 9.15 mA (since the current in the circuit is split evenly between the two branches, and IS1U1B1TS1U1B2=3.3 mA, we can divide the remaining current between IS1U1B1TS1U1B1 and IS1U1B1TS1U1B3, giving us 9.15 mA for each).

Figure 6 ’1 \text { Figure } 6 \text { '1 } Figure 6 ’1 -If R1=4

$\text { Figure } 6 \text { '1 }$ -If R₁=4