Question

In the circuit shown, each resistance is 2 $\Omega$. The potential $V_1$ (in volt) as indicated in the circuit, is equal to:

Answer 1

-9

Answer 2

The circuit can be analyzed using Kirchhoff's Current Law (KCL) at various nodes. Let's assign potentials to the key nodes.

1. Let the potential at the top arrow (which is usually considered ground in such problems unless specified otherwise) be $V_G = 0 \, V$.
2. Let the potential at the central junction where three resistors meet be $V_X$.
3. Let the potential at the node between the 2V battery and the first resistor in that branch be $V_A$.
4. Let the potential at the node between the two resistors in the 2V battery branch be $V_B$.
5. Let the potential at the node between the 9V battery and the resistor connecting to $V_X$ be $V_C$.
6. The potential $V_1$ is at the bottom left. Let's use it as $V_1$.
7. Let the potential at the node between $V_1$ and the 9V battery be $V_D$.

All resistances are $R = 2 \, \Omega$.

From the diagram, we can establish the following relationships:

• The positive terminal of the 2V battery is connected to node $V_A$. The negative terminal is at $V_1$. So, $V_A = V_1 + 2$.
• The positive terminal of the 9V battery is connected to node $V_C$. The negative terminal is at $V_D$. So, $V_C = V_D + 9$.

Now, let's write KCL equations for the nodes $V_X$, $V_B$, and $V_D$. Assume currents leaving a node are positive.

KCL at Node $V_X$:

The resistors connected to $V_X$ are:

• One to ground ($V_G = 0V$).
• One to $V_B$.
• One to $V_C$.

The sum of currents leaving $V_X$ is zero:

$$\frac{V_X - V_G}{R} + \frac{V_X - V_B}{R} + \frac{V_X - V_C}{R} = 0$$

Substitute $V_G = 0$ and $V_C = V_D + 9$:

$$\frac{V_X - 0}{R} + \frac{V_X - V_B}{R} + \frac{V_X - (V_D + 9)}{R} = 0$$

Multiply by $R$:

$$V_X + V_X - V_B + V_X - V_D - 9 = 0$$ $$3V_X - V_B - V_D = 9 \quad \text{(Equation 1)}$$

KCL at Node $V_B$:

The resistors connected to $V_B$ are:

• One to $V_X$.
• One to $V_A$.

The sum of currents leaving $V_B$ is zero:

$$\frac{V_B - V_X}{R} + \frac{V_B - V_A}{R} = 0$$

Substitute $V_A = V_1 + 2$:

$$\frac{V_B - V_X}{R} + \frac{V_B - (V_1 + 2)}{R} = 0$$

Multiply by $R$:

$$V_B - V_X + V_B - V_1 - 2 = 0$$ $$2V_B - V_X - V_1 = 2 \quad \text{(Equation 2)}$$

Current relationship in the 9V battery branch:

The current flowing through the resistor between $V_C$ and $V_X$ is the same as the current flowing through the 9V battery and the resistor between $V_D$ and $V_1$.

Let $I$ be the current flowing from $V_C$ to $V_X$.

$$I = \frac{V_C - V_X}{R}$$

This same current $I$ flows from $V_D$ to $V_1$ through the resistor $R_{D-1}$.

$$I = \frac{V_D - V_1}{R}$$

Therefore,

$$\frac{V_C - V_X}{R} = \frac{V_D - V_1}{R}$$ $$V_C - V_X = V_D - V_1$$

Substitute $V_C = V_D + 9$:

$$(V_D + 9) - V_X = V_D - V_1$$ $$V_D + 9 - V_X = V_D - V_1$$ $$9 - V_X = -V_1$$ $$V_X - V_1 = 9 \quad \text{(Equation 3)}$$

Now we have a system of three linear equations with three unknowns ($V_1, V_B, V_X, V_D$). We need to eliminate $V_B$ and $V_D$ to find $V_1$.

From Equation 3, express $V_X$ in terms of $V_1$:

$$V_X = V_1 + 9$$

Substitute $V_X$ into Equation 2:

$$2V_B - (V_1 + 9) - V_1 = 2$$ $$2V_B - 2V_1 - 9 = 2$$ $$2V_B = 2V_1 + 11$$ $$V_B = V_1 + \frac{11}{2} \quad \text{(Equation 4)}$$

Substitute $V_X$ into Equation 1:

$$3(V_1 + 9) - V_B - V_D = 9$$ $$3V_1 + 27 - V_B - V_D = 9$$ $$3V_1 - V_B - V_D = 9 - 27$$ $$3V_1 - V_B - V_D = -18 \quad \text{(Equation 5)}$$

Substitute $V_B$ from Equation 4 into Equation 5:

$$3V_1 - \left(V_1 + \frac{11}{2}\right) - V_D = -18$$ $$3V_1 - V_1 - \frac{11}{2} - V_D = -18$$ $$2V_1 - V_D = -18 + \frac{11}{2}$$ $$2V_1 - V_D = \frac{-36 + 11}{2}$$ $$2V_1 - V_D = -\frac{25}{2}$$ $$V_D = 2V_1 + \frac{25}{2} \quad \text{(Equation 6)}$$

Now, consider the current flowing through the 2V battery branch.

The current flowing from $V_A$ to $V_B$ is $I_{A-B} = (V_A - V_B)/R$.

This is the same current that flows through the 2V battery. So, $I_{1-A} = I_{A-B}$.

The current flowing from $V_1$ to $V_A$ through the battery means $I_{1-A} = (V_1 - V_A + 2)/R$ if we consider the battery as a voltage source with no internal resistance.

However, we have already used the potential relationships $V_A = V_1 + 2$.

The current from $V_1$ to $V_A$ is not directly through a resistor.

The current flowing out of $V_1$ goes into the 2V battery and into the resistor $R_{1-D}$.

Let's apply KCL at $V_1$.

Current leaving $V_1$ towards the 2V battery: Let this be $I_{1 \to \text{battery}}$.

Current leaving $V_1$ towards $V_D$: $(V_1 - V_D)/R$.

Sum of currents leaving $V_1$ must be zero.

$I_{1 \to \text{battery}} + \frac{V_1 - V_D}{R} = 0$.

Let's find $I_{1 \to \text{battery}}$. This current flows from $V_A$ through the resistor $R_{A-B}$ to $V_B$, and then from $V_B$ through $R_{B-X}$ to $V_X$.

So, $I_{1 \to \text{battery}} = I_{A-B} = \frac{V_A - V_B}{R}$.

$$I_{1 \to \text{battery}} = \frac{(V_1 + 2) - V_B}{R}$$

Substitute $V_B = V_1 + \frac{11}{2}$:

$$I_{1 \to \text{battery}} = \frac{V_1 + 2 - (V_1 + \frac{11}{2})}{R} = \frac{2 - \frac{11}{2}}{R} = \frac{\frac{4-11}{2}}{R} = \frac{-7/2}{R} = -\frac{7}{2R}$$

This means the current $I_{1 \to \text{battery}}$ is actually flowing into $V_1$ from the battery branch.

Now, apply KCL at $V_1$:

Current flowing into $V_1$ from the 2V battery branch is $7/(2R)$.

Current flowing out of $V_1$ into the resistor $R_{1-D}$ is $(V_1 - V_D)/R$.

So,

$$\frac{7}{2R} + \frac{V_1 - V_D}{R} = 0$$

Multiply by $R$:

$$\frac{7}{2} + V_1 - V_D = 0$$ $$V_D = V_1 + \frac{7}{2} \quad \text{(Equation 7)}$$

Now we have two expressions for $V_D$ (Equation 6 and Equation 7). Let's equate them:

$$2V_1 + \frac{25}{2} = V_1 + \frac{7}{2}$$ $$2V_1 - V_1 = \frac{7}{2} - \frac{25}{2}$$ $$V_1 = \frac{7 - 25}{2}$$ $$V_1 = \frac{-18}{2}$$ $$V_1 = -9 \, V$$

The potential $V_1$ is -9 Volt.