This circuit appears to show two voltage sources $E_1$ and $E_2$, a resistor $R$, and a potential difference $U_{ab}$ between points $a$ and $b$. The goal is to calculate the current $I$ through the resistor $R$.

Given dataThis circuit appears to show two voltage sources $E_1$ and $E_2$, a resistor $R$, and a potential difference $U_{ab}$ between points $a$ and $b$. The goal is to calculate the current $I$ through the resistor $R$.

Given

• $U_{ab} = 100 \, \text{V}$ (potential difference between points $a$ and $b$),
• $E_1 = 20 \, \text{V}$ (voltage source 1),
• $E_2 = 50 \, \text{V}$ (voltage source 2),
• $R = 10 \, \Omega$ (resistor value),
• We need to find the current $I$.

Approach:

Using Kirchhoff's Voltage Law (KVL), we can write the equation for the total voltage in the loop. From the diagram, we assume the following:

1. The current flows clockwise, starting from point $a$ to $b$ through the resistor.
2. The potential difference $U_{ab}$ is given as $100 \, \text{V}$.
3. The voltage drops across $E_1$ and $E_2$ need to be considered.

KVL Equation:

$U_{ab} = E_1 + E_2 + I \cdot R$ Now, let's solve for the current $I$:

$I = \frac{U_{ab} - (E_1 + E_2)}{R}$

Substituting the values:

$I = \frac{100 \, \text{V} - (20 \, \text{V} + 50 \, \text{V})}{10 \, \Omega}$ $I = \frac{100 \, \text{V} - 70 \, \text{V}}{10 \, \Omega}$ $I = \frac{30 \, \text{V}}{10 \, \Omega} = 3 \, \text{A}$

Thus, the current $I$ is 3 A.

Would you like a detailed breakdown of each step or further clarification?

Here are 5 related questions:

1. How does Kirchhoff's Voltage Law apply to different circuits?
2. What happens to the current if one of the voltage sources is reversed?
3. How does changing the resistance $R$ affect the current in the circuit?
4. How can we determine the power dissipated by the resistor $R$?
5. How would the result change if the voltage difference $U_{ab}$ was smaller?

Tip: In circuits with multiple sources, always check the direction of each source when applying Kirchhoff's Laws to avoid sign errors.