For the diagram shown in the figure, the following is true for contour 'C':

To determine the correct equation for contour 'C', we need to apply Kirchhoff's Voltage Law (KVL) to that specific loop. KVL states that the sum of the voltage drops around any closed loop in a circuit must equal the sum of the voltage rises in that loop.

Let's analyze contour 'C':

• We start from point C and move clockwise.
• We encounter the voltage source E₃. Since we are moving in the same direction as the positive terminal of E₃ (assuming the longer bar indicates the positive terminal), this is a voltage rise. So we add +E₃.
• Next, we encounter resistor R₅. The current flowing through R₅ is I₅. We are moving in the same direction as the assumed current I₅. Therefore, this is a voltage drop across R₅, which is I₅R₅. So we subtract -I₅R₅.
• Then, we encounter resistor R₂. The current flowing through R₂ is I₂. We are moving in the opposite direction of the assumed current I₂. Therefore, this is a voltage rise across R₂, which is -I₂R₂. So we add -I₂R₂.
• Finally, we encounter the voltage source E₂. We are moving in the opposite direction of the positive terminal of E₂. Therefore, this is a voltage drop across E₂, which is -E₂. So we subtract -E₂.

Applying KVL for contour 'C' (clockwise):

\[ -E_2 - I_2R_2 + I_5R_5 + E_3 = 0 \]

Rearranging the terms to match the options:

\[ -E_2 + E_3 = I_2R_2 - I_5R_5 \]

Let's re-examine the direction of current I₂ and our path through R₂.

Assuming we traverse the loop C clockwise:

• Voltage source E₃: +E₃ (moving from - to +)
• Resistor R₅: The current I₅ flows downwards. We are moving downwards through R₅, so it's a voltage drop: -I₅R₅.
• Resistor R₂: The current I₂ flows from left to right. We are moving from right to left through R₂, so it's a voltage rise: +I₂R₂.
• Voltage source E₂: We are moving from + to - through E₂, so it's a voltage drop: -E₂.

Summing these up:

\[ E_3 - I_5R_5 + I_2R_2 - E_2 = 0 \]

Rearranging to match the options:

\[ -E_2 + E_3 = I_5R_5 - I_2R_2 \]

Let's check the provided options against this derivation. It seems there might be a discrepancy in the assumed direction of I₂ or how it relates to the loop.

Let's re-evaluate the currents and loop directions carefully.

Contour C is the loop formed by E₂, R₂, and R₅. Let's assume clockwise traversal for contour C.

• Starting from E₂, moving from positive to negative terminal: -E₂
• Moving through R₂ in the direction opposite to I₂: +I₂R₂
• Moving through R₅ in the direction of I₅: -I₅R₅
• Moving through E₃ from negative to positive terminal: +E₃

So, the equation for contour C is:

\[ -E_2 + I_2R_2 - I_5R_5 + E_3 = 0 \]

Rearranging this equation:

\[ -E_2 + E_3 = I_5R_5 - I_2R_2 \]

Let's re-examine the options. None of the options exactly match this. It's possible the direction of current I₂ in the diagram is intended to be from right to left, or the direction of traversal of the loop is different.

Let's assume the options are correct and try to match one. The third option is -E₂ + E₃ = -I₂R₂ + I₅R₅. This would imply that when traversing the loop, we have a voltage rise of -E₂ + E₃, and voltage drops of I₂R₂ and -I₅R₅ (which is a rise).

Let's assume clockwise traversal of contour C and the given currents.

• Voltage source E₃: +E₃ (moving from - to +)
• Resistor R₅: Current I₅ is downwards. We traverse R₅ downwards. Voltage drop is -I₅R₅.
• Resistor R₂: Current I₂ is from left to right. We traverse R₂ from right to left. Voltage rise is +I₂R₂.
• Voltage source E₂: Moving from + to - is a voltage drop -E₂.

Summing up: +E₃ - I₅R₅ + I₂R₂ - E₂ = 0

Rearranging: -E₂ + E₃ = I₅R₅ - I₂R₂

This still does not match any option. Let's reconsider the directions and options.

Let's look at the third option again: -E₂ + E₃ = -I₂R₂ + I₅R₅.

If this is correct, then when traversing the loop C:

• Voltage rises are E₂ and -E₃ OR voltage drops are -E₂ and +E₃.
• Voltage drops are I₂R₂ and -I₅R₅.

Let's assume clockwise traversal of contour C:

• +E₃ (rise)
• -I₅R₅ (drop)
• +I₂R₂ (rise, if I₂ is from right to left, but it's shown from left to right). If I₂ is from left to right, then traversing from right to left is a rise. So +I₂R₂.
• -E₂ (drop)

So, E₃ - I₅R₅ + I₂R₂ - E₂ = 0

This leads to -E₂ + E₃ = I₅R₅ - I₂R₂.

Let's assume the intended answer is the third option: -E₂ + E₃ = -I₂R₂ + I₅R₅.

For this to be true, when traversing contour C clockwise:

• Voltage rise: +E₃
• Voltage drop: -I₅R₅
• Voltage rise: -I₂R₂ (This means the current I₂ is flowing in the same direction as our traversal of R₂, which is from left to right. So, if we traverse clockwise, we go from right to left. If I₂ is from left to right, then traversing R₂ from right to left is a voltage rise: +I₂R₂. This contradicts the option.)
• Voltage drop: +E₂

Let's try counter-clockwise traversal of contour C:

• Voltage source E₂: +E₂ (moving from - to +)
• Resistor R₂: Current I₂ is from left to right. We traverse R₂ from left to right. Voltage drop is -I₂R₂.
• Resistor R₅: Current I₅ is downwards. We traverse R₅ upwards. Voltage rise is +I₅R₅.
• Voltage source E₃: Moving from + to - is a voltage drop -E₃.

Summing up: +E₂ - I₂R₂ + I₅R₅ - E₃ = 0

Rearranging:

\[ E_2 - E_3 = I_2R_2 - I_5R_5 \]

This also does not match any option.

Let's re-examine the diagram and options carefully. The currents are labeled I₂ and I₅. The voltage sources are E₂ and E₃. The resistors are R₂ and R₅.

Consider the loop C (containing E₂, R₂, R₅, E₃). Let's assume a clockwise traversal starting from the bottom left corner of the loop (where E₃'s negative terminal is).

1. Through E₃, from - to +: +E₃
2. Through R₅, in the direction of I₅: -I₅R₅
3. Through R₂, in the direction opposite to I₂ (I₂ is shown left-to-right, our path is right-to-left): +I₂R₂
4. Through E₂, from + to -: -E₂

So, +E₃ - I₅R₅ + I₂R₂ - E₂ = 0

Rearranging: -E₂ + E₃ = I₅R₅ - I₂R₂

There seems to be a sign error in the options or my interpretation. Let's check the third option again: -E₂ + E₃ = -I₂R₂ + I₅R₅.

If we assume the current I₂ flows from right to left, then traversing R₂ from right to left would be a voltage drop of -I₂R₂. And if current I₅ flows upwards, then traversing R₅ upwards would be a voltage rise of +I₅R₅.

However, the diagram clearly shows I₂ flowing from left to right and I₅ flowing downwards.

Let's assume the option -E₂ + E₃ = -I₂R₂ + I₅R₅ is correct and see if we can justify it.

If we assume a counter-clockwise traversal for contour C:

• Through E₂, from - to +: +E₂
• Through R₂, in the direction of I₂: -I₂R₂
• Through R₅, in the direction opposite to I₅ (I₅ is downwards, our path is upwards): +I₅R₅
• Through E₃, from + to -: -E₃

Summing up: +E₂ - I₂R₂ + I₅R₅ - E₃ = 0

Rearranging: E₂ - E₃ = I₂R₂ - I₅R₅

Still not matching. Let's revisit the third option: -E₂ + E₃ = -I₂R₂ + I₅R₅.

This equation can be rewritten as: E₂ - E₃ = I₂R₂ - I₅R₅.

This is exactly what we derived from a counter-clockwise traversal of contour C.

Therefore, the correct answer is indeed the one that corresponds to this equation.

Checking the options:

• -E₂ + E₃ = I₂R₂ - I₅R₅ (Option 1)
• E₂ + E₃ = -I₂R₂ + I₅R₅ (Option 2)
• -E₂ + E₃ = -I₂R₂ + I₅R₅ (Option 3)
• -E₂ + E₃ = I₂R₂ + I₅R₅ (Option 4)

The equation we derived from counter-clockwise traversal is E₂ - E₃ = I₂R₂ - I₅R₅. If we multiply this by -1, we get -E₂ + E₃ = -I₂R₂ + I₅R₅. This matches option 3.

Thus, the equation for contour 'C' (using Kirchhoff's Voltage Law) with counter-clockwise traversal is:

\[ E_2 - I_2R_2 + I_5R_5 - E_3 = 0 \]

Rearranging this to match the third option:

\[ -E_2 + E_3 = -I_2R_2 + I_5R_5 \]

Therefore, the third option is the correct one.

Final Answer Derivation:

Applying Kirchhoff's Voltage Law to contour 'C' with a counter-clockwise traversal:

1. Start at the negative terminal of E₂ and move towards the positive terminal: voltage rise of +E₂.
2. Move through resistor R₂ in the same direction as current I₂ (left to right): voltage drop of -I₂R₂.
3. Move through resistor R₅ in the direction opposite to current I₅ (I₅ is downwards, we are moving upwards): voltage rise of +I₅R₅.
4. Move through voltage source E₃ from the positive to the negative terminal: voltage drop of -E₃.

Summing these voltages around the loop:

\[ E_2 - I_2R_2 + I_5R_5 - E_3 = 0 \]

Rearranging the equation to match the format of the options:

\[ E_2 - E_3 = I_2R_2 - I_5R_5 \]

Multiplying the entire equation by -1 to get the form -E₂ + E₃:

\[ -(E_2 - E_3) = -(I_2R_2 - I_5R_5) \]

\[ -E_2 + E_3 = -I_2R_2 + I_5R_5 \]

This matches the third option.