A circuit consists of 5 resistors and 3 batteries (see diagram); the connecting wires have all a negligible resistance. The values for R1 , R2 , R3 , R4 , and R5 are 14 Ω, 41 Ω, 68 Ω, 83 Ω, and 110 Ωrespectively. The batteries have a negligible internal resistance; their voltages , V1 , V2 and V3 , are 12 V, 23 V and 50 V, respectively (for their polarities, see the diagram).

(a)

Is the current I1 flowing in the direction shown? Choose which direction I1 is really flowing below:
What is it magnitude in Amperes?

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Is the current I2 flowing in the direction shown? Choose which direction I2 is really flowing below:
What is it magnitude in Amperes?

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Is the current I4 flowing in the direction shown? Choose which direction I4 is really flowing below:
What is it magnitude in Amperes?

unanswered

(b)

What is the potential difference (pay attention to the sign!) in Volts between the points A&P : VA-VP ?

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What is the potential difference (pay attention to the sign!) in Volts between the points P&N : VP-VN ?

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What is the potential difference (pay attention to the sign!) in Volts between the points G&D : VG-VD ?

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Please some one answer.

To determine the current flowing in each direction and the potential differences between different points in the circuit, we can use Kirchhoff's laws.

(a) Currents in the circuit:

1. To determine the direction of current I1:
- The positive terminal of battery V1 is connected to the positive terminal of battery V2 and resistor R1.
- Since the positive terminal is connected to the resistor, the current will flow from positive to negative direction through resistor R1.
- Therefore, the direction of current I1 is from right to left.

2. To determine the direction of current I2:
- The positive terminal of battery V2 is connected to the positive terminal of battery V3 and resistor R2.
- Since the positive terminal is connected to the resistor, the current will flow from positive to negative direction through resistor R2.
- Therefore, the direction of current I2 is from left to right.

3. To determine the direction of current I4:
- The positive terminal of battery V2 is connected to resistor R4, and then to resistor R5.
- Since the positive terminal is connected to resistor R4, the current will flow from positive to negative direction through resistor R4.
- Therefore, the direction of current I4 is from right to left.

Now let's calculate the magnitudes of these currents:

To calculate the current I1:
- As per Ohm's Law, V = IR, where V is the potential difference across the resistor and R is the resistance.
- The potential difference V1 across resistor R1 is 12 V (given).
- Substituting the values, we have: 12 V = I1 * 14 Ω.
- Solving for I1, we get: I1 = 12 V / 14 Ω.

To calculate the current I2:
- The potential difference V2 across resistor R2 is 23 V (given).
- Substituting the values, we have: 23 V = I2 * 41 Ω.
- Solving for I2, we get: I2 = 23 V / 41 Ω.

To calculate the current I4:
- The potential difference V2 across resistor R4 is 23 V (given).
- Substituting the values, we have: 23 V = I4 * 83 Ω.
- Solving for I4, we get: I4 = 23 V / 83 Ω.

(b) Potential differences in the circuit:

1. Potential difference between points A and P (VA-VP):
- The potential difference VA at point A is given as 0 V (reference point).
- The potential difference VP can be calculated as follows:
- VA - VP = 0 V - VP = -VP.
- Therefore, the potential difference between points A and P is -VP V.

2. Potential difference between points P and N (VP-VN):
- The potential difference VP can be calculated as follows:
- VP - VN = VP - 0 V = VP V.
- Therefore, the potential difference between points P and N is VP V.

3. Potential difference between points G and D (VG-VD):
- The potential difference VG can be calculated as follows:
- VG - VD = 50 V - VD = VG - 0 V = VG V.
- Therefore, the potential difference between points G and D is VG V.

To solve this circuit problem, we'll need to apply Ohm's Law and Kirchhoff's laws.

(a)

To determine the direction of the currents (I1, I2, I4) and their magnitudes, we can use Kirchhoff's current law (KCL). KCL states that the sum of currents entering a node (junction) equals the sum of currents leaving that node.

From the diagram, we can see that the current I1 enters the node, and there are no other paths for it to leave. Therefore, I1 must be flowing in the direction shown. Similarly, I2 and I4 also flow in the directions shown since there are no other paths for them to leave.

To find their magnitudes, we'll use Ohm's Law. Ohm's Law states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R). Mathematically, I = V/R.

For I1:
V1 = I1 * R1
12V = I1 * 14Ω
I1 = 12V / 14Ω ≈ 0.857 A (rounded to three decimal places)

For I2:
V2 = I2 * R2
23V = I2 * 41Ω
I2 = 23V / 41Ω ≈ 0.561 A (rounded to three decimal places)

For I4:
V3 = I4 * R5
50V = I4 * 110Ω
I4 = 50V / 110Ω ≈ 0.455 A (rounded to three decimal places)

Therefore, the magnitudes of the currents are:
I1 ≈ 0.857 A
I2 ≈ 0.561 A
I4 ≈ 0.455 A

(b)

To find the potential differences between the given points, we can use Kirchhoff's voltage law (KVL). KVL states that the sum of potential differences in a closed loop in a circuit is zero.

For the potential difference between points A and P (VA-VP), we can consider the loop APGA. Since there are no resistors or batteries in this loop, the potential difference VA-VP is zero.

For the potential difference between points P and N (VP-VN), we can consider the loop PNBVP. The potential differences across the batteries are V2 = 23V and V3 = 50V, and they oppose each other as shown in the diagram. Therefore, VP-VN = V2 - V3 = 23V - 50V = -27V (negative sign indicates the direction of the potential difference).

For the potential difference between points G and D (VG-VD), we can consider the loop DGCD. This loop includes resistors R4 and R5. So, VG-VD = (I4 * R5) - (I2 * R4) = (0.455A * 110Ω) - (0.561A * 83Ω) = 49.9V - 46.563V ≈ 3.337V (rounded to three decimal places). The positive sign indicates the direction of the potential difference.

Therefore, the potential differences are:
VA-VP = 0V
VP-VN = -27V
VG-VD ≈ 3.337V