Question

for each circuit below, find all missing resistances, voltages, and currents.
1.
2.
3.
4.

Explanation:

Step1: Recall Ohm's law

Ohm's law is $V = IR$, where $V$ is voltage, $I$ is current, and $R$ is resistance. For series - connected resistors, the equivalent resistance $R_{eq}=\sum_{i = 1}^{n}R_{i}$.

Circuit 1:

Step2: Calculate equivalent resistance

The two resistors are in series. So, $R_{eq}=1\Omega + 1\Omega=2\Omega$.

Step3: Calculate current

Using Ohm's law $I=\frac{V}{R_{eq}}$. Given $V = 6V$ and $R_{eq}=2\Omega$, then $I=\frac{6V}{2\Omega}=3A$.

Step4: Calculate voltage across each resistor

Using $V = IR$, for the first $1\Omega$ resistor, $V_1=I\times R_1=3A\times1\Omega = 3V$. For the second $1\Omega$ resistor, $V_2=I\times R_2=3A\times1\Omega = 3V$.

Circuit 2:

Step5: Calculate equivalent resistance

The two resistors are in series. $R_{eq}=1\Omega+2\Omega = 3\Omega$.

Step6: Calculate current

Using Ohm's law $I=\frac{V}{R_{eq}}$. Given $V = 6V$ and $R_{eq}=3\Omega$, then $I=\frac{6V}{3\Omega}=2A$.

Step7: Calculate voltage across each resistor

For the $1\Omega$ resistor, $V_1=I\times R_1=2A\times1\Omega = 2V$. For the $2\Omega$ resistor, $V_2=I\times R_2=2A\times2\Omega = 4V$.

Circuit 3:

Step8: Calculate equivalent voltage

The two batteries are in series, so $V_{eq}=1.5V + 1.5V=3V$.

Step9: Calculate equivalent resistance

Let the equivalent resistance be $R_{eq}$. Using Ohm's law $R_{eq}=\frac{V_{eq}}{I}$. Given $V_{eq}=3V$ and $I = 0.5A$, then $R_{eq}=\frac{3V}{0.5A}=6\Omega$. Let the three resistors be $R_1$, $R_2$, $R_3$. Since $R_{eq}=R_1 + R_2+R_3$, and we know $I = 0.5A$, we can find the voltage across each resistor. If we assume the resistors are $R_1$, $R_2$, $R_3$, $V_1=I\times R_1$, $V_2=I\times R_2$, $V_3=I\times R_3$. But since the individual resistances are not given separately, if we assume they are equal (for the sake of finding voltage across each in a symmetric case), $R_1=R_2=R_3 = 2\Omega$ (because $R_{eq}=6\Omega$), then $V_1=0.5A\times2\Omega = 1V$, $V_2=0.5A\times2\Omega = 1V$, $V_3=0.5A\times2\Omega = 1V$.

Circuit 4:

Step10: Calculate equivalent resistance

The three resistors are in series. $R_{eq}=2\Omega+3\Omega + 1\Omega=6\Omega$.

Step11: Calculate current

Using Ohm's law $I=\frac{V}{R_{eq}}$. Given $V = 9V$ and $R_{eq}=6\Omega$, then $I=\frac{9V}{6\Omega}=1.5A$.

Step12: Calculate voltage across each resistor

For the $2\Omega$ resistor, $V_1=I\times R_1=1.5A\times2\Omega = 3V$. For the $3\Omega$ resistor, $V_2=I\times R_2=1.5A\times3\Omega = 4.5V$. For the $1\Omega$ resistor, $V_3=I\times R_3=1.5A\times1\Omega = 1.5V$.

Answer:

Circuit 1: $R_{eq}=2\Omega$, $I = 3A$, $V_1 = 3V$, $V_2 = 3V$
Circuit 2: $R_{eq}=3\Omega$, $I = 2A$, $V_1 = 2V$, $V_2 = 4V$
Circuit 3: $V_{eq}=3V$, $R_{eq}=6\Omega$ (assuming equal - valued resistors of $2\Omega$ each), $V_1 = 1V$, $V_2 = 1V$, $V_3 = 1V$
Circuit 4: $R_{eq}=6\Omega$, $I = 1.5A$, $V_1 = 3V$, $V_2 = 4.5V$, $V_3 = 1.5V$