Question

In the circuit shown in the figure (Figure 1) both batteries have insignificant internal resistance and the idealized ammeter reads 1.30 A in the direction shown. er reads 1.30nal resistan gure 1) both Part A Find the erf of the battery. 10 AEGO ? E = Figure Submit Request Answer 1 of 1 Part B Is the polarity shown correct? 12.0 12 WW + 8=? yes 48. 03 no 75.0 VT 3 15.0 25 T?

Answer 1

The emf of the battery is 13.00 V and the polarity shown in the circuit is correct.

Step-by-step explanation:

In the given circuit, the idealized ammeter reads a current of 1.30 A in the direction shown. To find the emf of the battery, we can apply Kirchhoff's voltage law (KVL) around the loop.

The sum of the voltage drops across the resistors must equal the emf of the battery:

emf - 10.00Ω x 1.30 A = 0

Solving for emf, we get:

emf = 13.00 V

Therefore, the emf of the battery is 13.00 V.

To determine if the polarity shown is correct, we need to look at the direction of the ammeter reading and compare it with the direction of the battery symbol in the circuit diagram.

Since the idealized ammeter reads a positive current of 1.30 A in the direction shown, the polarity of the battery is correct.

Answer 2

The emf of the batteries in the circuit is 26.00 V and the polarity shown is correct.

Step-by-step explanation:

The circuit shown in the figure consists of two batteries connected in series with an external resistor. The ammeter reads a current of 1.30 A in the direction shown. To find the emf of the batteries, we can use Ohm's Law. Ohm's Law states that V = IR, where V is the voltage, I is the current, and R is the resistance.

In this case, the current through the external resistor is 1.30 A and the resistance is 10.00 Ω. Therefore, the voltage across the resistor is V = (1.30 A)(10.00 Ω) = 13.00 V.

Since the two batteries are connected in series, the total emf of the circuit is equal to the sum of the individual emfs. Therefore, the emf of the batteries can be calculated as:

Emf = 13.00 V + 13.00 V = 26.00 V

As for the polarity of the batteries, the positive terminal of one battery is connected to the negative terminal of the other battery. This ensures that the emfs of the batteries are additive. Therefore, the polarity shown in the figure is correct.