9. the voltage at point 1 equals \n a 2 v \n b 6 v \n c 9 v \n d 10 v \n 15 v\nr1 2 kω\nr2 3 kω\nrx 10…

9. the voltage at point 1 equals \n a 2 v \n b 6 v \n c 9 v \n d 10 v \n 15 v\nr1 2 kω\nr2 3 kω\nrx 10 kω\nrs 5 kω\n1\n2

9. the voltage at point 1 equals \n a 2 v \n b 6 v \n c 9 v \n d 10 v \n 15 v\nr1 2 kω\nr2 3 kω\nrx 10 kω\nrs 5 kω\n1\n2

Answer

Explanation:

Step1: Calculate total resistance in series part

Resistors $R1 = 2k\Omega$ and $R2=3k\Omega$ are in series. The total resistance $R_{total1}=R1 + R2=2k\Omega+3k\Omega = 5k\Omega$.

Step2: Use voltage - division rule

The voltage - division rule for two - resistors in series is $V_x=\frac{R_x}{R_{total}}V_{source}$. Here, the source voltage $V = 15V$, and we want to find the voltage across $R2$. $V_1=\frac{R2}{R1 + R2}\times V=\frac{3k\Omega}{2k\Omega+3k\Omega}\times15V$. $V_1=\frac{3}{5}\times15V = 9V$.

Answer:

C. 9 V