problem 17: (6% of assignment value) consider the circuit shown, where r1 = 10 kω, r2 = 24 kω, r3 = 16 kω…

Explanation:

Step1: Find total resistance

The resistors are in series. The total resistance $R_{total}=R_1 + R_2+R_3+R_4+R_5$. Given $R_1 = 10\ k\Omega$, $R_2 = 24\ k\Omega$, $R_3 = 16\ k\Omega$, $R_4 = 10\ k\Omega$, $R_5 = 18\ k\Omega$. So $R_{total}=(10 + 24+16+10+18)\ k\Omega=78\ k\Omega$.

Step2: Find total voltage

The total voltage $V = V_1+V_2$, where $V_1 = 12.0\ V$ and $V_2 = 24.0\ V$. So $V=(12 + 24)\ V=36\ V$.

Step3: Find the current in the circuit

Using Ohm's law $I=\frac{V}{R}$, where $V$ is the total voltage and $R$ is the total resistance. So $I=\frac{36\ V}{78\times10^{3}\ \Omega}\approx0.462\times10^{- 3}\ A$.

Step4: Find the potential - difference across $R_3$

Using Ohm's law $V_3 = I\times R_3$. Substitute $I = 0.462\times10^{-3}\ A$ and $R_3 = 16\times10^{3}\ \Omega$. So $V_3=(0.462\times10^{-3}\ A)\times(16\times10^{3}\ \Omega)=7.39\ V$.

Answer:

$7.39$