Internal resistance and electromotive force problem

[nutzweb]

help me with this please... i just couldn't get it.

1) a complete circuit consists of a 12.0 V battery with a 4.50 ohm resistor and switch. the internal resistance of the battery is 0.30 ohms when switch is open. what does the voltmeter read when placed:

a. across the terminal of the battery when the switch is open
b. across the resistor when the switch is open
c. across the terminal of the battery when the switch is closed
d. across the resistor when the switch is closed

2) when the switch S is open, the voltmeter V reads 2.0 V. when the switch is closed, the voltmeter reading drops to 1.50 V and the ammeter reads 1.20 A. find the emf (electromotive force) and the internal resistance of the battery. assume that the two meters are ideal so they don't affect the circuit.

 

[Andrew Mason]

help me with this please... i just couldn't get it.

1) a complete circuit consists of a 12.0 V battery with a 4.50 ohm resistor and switch. the internal resistance of the battery is 0.30 ohms when switch is open. what does the voltmeter read when placed:

a. across the terminal of the battery when the switch is open
b. across the resistor when the switch is open
c. across the terminal of the battery when the switch is closed
d. across the resistor when the switch is closed

2) when the switch S is open, the voltmeter V reads 2.0 V. when the switch is closed, the voltmeter reading drops to 1.50 V and the ammeter reads 1.20 A. find the emf (electromotive force) and the internal resistance of the battery. assume that the two meters are ideal so they don't affect the circuit.

V=IR. Ohm's law.

AM

 

[thepatientmental]

First, let's define internal resistance and electromotive force (emf). Internal resistance is the resistance within the battery itself, which can affect the voltage output of the battery. Electromotive force is the voltage produced by the battery, which is equal to the potential difference between the positive and negative terminals.

Now, let's tackle the first question. When the switch is open, it means that there is no current flowing in the circuit. This means that the voltmeter will read the full emf of the battery, which in this case is 12.0 V. This is because there is no voltage drop across the internal resistance since there is no current flowing through it. Therefore, the voltmeter will read 12.0 V when placed across the terminals of the battery.

When the switch is open, the voltmeter will read the voltage drop across the resistor, which is calculated using Ohm's Law (V = IR). Plugging in the values given, the voltmeter will read 1.2 V (0.3 ohms x 4.5 ohms = 1.2 V).

Now, let's move on to the second question. We are given the voltmeter reading and ammeter reading when the switch is open and closed. When the switch is open, the voltmeter reads the full emf of the battery, which we can calculate by rearranging Ohm's Law (V = IR) to solve for emf. Plugging in the values given, we get an emf of 2.0 V.

When the switch is closed, the voltmeter reading drops to 1.5 V. This is because there is now a voltage drop across both the internal resistance and the resistor. Using Kirchhoff's Voltage Law (KVL), we can set up an equation: emf - IR - IR = 1.5 V. Plugging in the values and solving for emf, we get an emf of 2.5 V.

Using the two values of emf (2.0 V and 2.5 V), we can solve for the internal resistance of the battery. We can set up another equation using Kirchhoff's Voltage Law: emf - IR = V. Plugging in the values and solving for R, we get an internal resistance of 0.5 ohms.

In conclusion, internal resistance and electromotive force can affect the voltage output