Photo AI

In the circuit diagram below, resistor R, with a resistance of 5.6 Ω, is connected, together with a switch, an ammeter and a high-resistance voltmeter, to a battery with an unknown internal resistance, r - NSC Physical Sciences - Question 8 - 2019 - Paper 1

Question icon

Question 8

In-the-circuit-diagram-below,-resistor-R,-with-a-resistance-of-5.6-Ω,-is-connected,-together-with-a-switch,-an-ammeter-and-a-high-resistance-voltmeter,-to-a-battery-with-an-unknown-internal-resistance,-r-NSC Physical Sciences-Question 8-2019-Paper 1.png

In the circuit diagram below, resistor R, with a resistance of 5.6 Ω, is connected, together with a switch, an ammeter and a high-resistance voltmeter, to a battery ... show full transcript

Worked Solution & Example Answer:In the circuit diagram below, resistor R, with a resistance of 5.6 Ω, is connected, together with a switch, an ammeter and a high-resistance voltmeter, to a battery with an unknown internal resistance, r - NSC Physical Sciences - Question 8 - 2019 - Paper 1

Step 1

Define the term emf of a battery.

96%

114 rated

Answer

The emf (electromotive force) of a battery is defined as the maximum potential difference provided by the battery when no current is flowing. It represents the energy supplied by the battery per coulomb of charge passing through it.

Step 2

Write down the value of the emf of the battery.

99%

104 rated

Answer

The value of the emf of the battery is 13 V.

Step 3

When switch S is CLOSED, calculate the current through resistor R.

96%

101 rated

Answer

Using Ohm's Law, we can determine the current (I) through resistor R:

I=VR=10.5V5.6Ω1.88AI = \frac{V}{R} = \frac{10.5 V}{5.6 \Omega} \approx 1.88 A

Step 4

When switch S is CLOSED, calculate the power dissipated in resistor R.

98%

120 rated

Answer

The power (P) dissipated in resistor R can be calculated using the formula:

P=I2R=(1.88A)2×5.6Ω19.688WP = I^2 R = (1.88 A)^2 \times 5.6 \Omega \approx 19.688 W

Step 5

When switch S is CLOSED, calculate the internal resistance, r, of the battery.

97%

117 rated

Answer

We can establish the relationship between voltage, current, and internal resistance:

ε=I(R+r)\n13=1.88(5.6+r)\nSolvingforr,wefindr1.31Ω.\varepsilon = I(R + r)\n13 = 1.88(5.6 + r)\nSolving for r, we find r \approx 1.31 \Omega.

Hence, the internal resistance of the battery is approximately 1.31 Ω.

Step 6

How would the voltmeter reading change? Choose from INCREASES, DECREASES or REMAINS THE SAME.

97%

121 rated

Answer

The voltmeter reading would DECREASE.

Step 7

Calculate resistance X.

96%

114 rated

Answer

Using the circuit with two identical resistors:

For two resistors in parallel:

\nUsing Ohm's law: $$I = \frac{V}{R_{total}};\n\n4 A = \frac{13 V}{2.8 \Omega}\n\nTherefore, X \approx 1.49 \Omega.$$

Join the NSC students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;