Question

the circuit shown has a 10.0 v battery connected to a cube with 2.00 mm sides and a resistivity 24 as well as two cylinders with radius 1.00 mm, length 1.00 cm and resistivity 1.80 . assuming the junction is grounded where the resistors meet, what is the potential at point a?

Answer 1

The potential at point A in the circuit is 36.96 V.

Step-by-step explanation:

In this circuit, the potential at point A can be found by calculating the voltage drop across the resistors.

First, let's calculate the potential drop across the cube resistor. The resistance of the cube is given as 24 Ω and the length of each side is 2 mm. Using the formula V = IR, where V is the potential drop, I is the current, and R is the resistance, we can find:

Vcube = (2 mm)^2 * 10 A/m * 24 Ω = 0.96 V

Next, let's calculate the potential drop across the cylinders. The resistance of each cylinder is given as 1.80 Ω and the radius is 1 mm. Using the same formula, we can find:

Vcylinder = (2 * π * 1 mm) * 10 A/m * 1.80 Ω = 36 V

Since the junction is grounded, the potential at point A is the sum of the potential drops: VA = Vcube + Vcylinder = 0.96 V + 36 V = 36.96 V