Physics, 29.07.2019 11:30 BreBreDoeCCx

In the following figure, what is the value of the potential at points a and b? (figure 1) picture a square. there is a 15 v battery on the left side, a 4ohm resistor on top, a 5v battery on the right side and a 1 ohm resistor on bottom. i need the potential at the upper right corner, and then the potential at the lower left corner.

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Answer from: ShevyPrice

(a) The potential at the upper right of the square is 13.32 V.

(b) The potential at the lower left corner of the square is 3.33 V.

Net voltage in the circuit

The net voltage in the circuit is determined by applying Kirchhoff rule assuming the current flow is anticlockwise.

∑V = 0

Assuming the following

Top of the square = positiveBottom of the square = negativeRight of the square = positiveLeft of the square = negative

-15 + (4I) + 5 - (1I) = 0

-10 + 3I = 0

3I = 10

I = 10/3 = 3.33 A

Potential at the upper right = 3.33 x 4 = 13.32 V

Potential at the lower left corner = 1 x 3.33 = 3.33 V.

Learn more about potential difference here: link

Answer from: kevinhill185

It depends on how the polarities of the two batteries are placed.
Here I assume that the battery on the left side has polarity- on top and polarity-+on bottom, while the battery on the right side has polarity- on bottom and polarity+ on top, so the current is flowing in anti-clockwise direction.

To solve the problem, we must first calculate the current flowing in the circuit. For this, we use Kirchoff's voltage law:
$\sum V_i =0$ (1)
We must consider both the emf of the batteries and the voltage drops on the resistors, which for sign convention we must take as negative. So (1) becomes
$15 V - (1 \Omega \cdot I)+5 V -(4 \Omega \cdot I)=0$
where I is the current flowing.
By solving the equation, we get:
20 V=5 I

So, assuming the potential at top left corner is 0 V (because it is at same potential of the negative polarity of the 15 V battery), then the voltage drop on the $4 \Omega$ resistor is $\Delta V = IR=(4 A)(4 \Omega)=16 V$ , and so the potential at the top right corner is 16 V. Instead, the bottom left corner is at same potential of the positive terminal of the 15 V battery, so at 15 V.

In case the polarities of the 5 V battery are inverted, the exercise must be re-done in the same way by changing the sign of the emf of the 5 V battery in the Kirchoff's law.

Answer from: Quest

force tends to be low and long diustance masses are high

Answer from: Quest

The answer is a i think

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In the following figure, what is the value of the potential at points a and b? (figure 1) picture...

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