In this experiment, you will verifi that the current entering a junction in a circuit equals the current leaving it. You connects three resistors as shown i...
Question 1 Report
In this experiment, you will verifi that the current entering a junction in a circuit equals the current leaving it. You connects three resistors as shown in Fig. 57.1: a 10 Ω resistor in series with a parallel combination of a 22 Ω and a 33 Ω resistor. You places ammeters at positions P, Q and R.
| Ammeter position | I / A |
|---|---|
| P (main circuit) | 0.25 |
| Q (22 Ω branch) | 0.15 |
| R (33 Ω branch) | 0.10 |
(a) Record the three ammeter readings in Table 57.1. [1]
(b) Calculate the sum of the currents in the two branches (Q + R). [1]
(c) Compare this sum with the reading at P. State your conclusion. [2]
(d) Calculate the p.d. across the 22 Ω resistor using V = I × R. [1]
(e) Calculate the p.d. across the 10 Ω resistor. [1]
(f) Show that V across 10 Ω + V across parallel pair approximately equals 6.0 V. [1]
(g) Explain why the current through the 22 Ω resistor is greater than through the 33 Ω resistor. [1]
(h) State one precaution you should take when moving the ammeter between positions. [1]
(i) State one source of error in this experiment. [1]
Answer Details
(a) The three ammeter readings are recorded: P = 0.25 A, Q = 0.15 A, R = 0.10 A. [1]
(b) The sum of the currents in the two parallel branches is:
\[ I_Q + I_R = 0.15 + 0.10 = 0.25 \text{ A} \]
[1]
(c) The sum of the branch currents (0.25 A) is equal to the reading at ammeter P (0.25 A). [1] This confirms Kirchhoff's first law (the junction rule): the total current entering a junction equals the total current leaving it. All the current from the battery passes through the 10 Ω resistor and then splits at the junction into two paths; the branch currents recombine to give the original total. [1]
(d) The potential difference across the 22 Ω resistor is found using \( V = I \times R \):
\[ V = 0.15 \times 22 = 3.3 \text{ V} \]
[1]
(e) The potential difference across the 10 Ω resistor is:
\[ V = I \times R = 0.25 \times 10 = 2.5 \text{ V} \]
The current through the 10 Ω resistor is the total circuit current (0.25 A), since it is in the main loop before the parallel junction. [1]
(f) The total voltage around the loop should equal the battery voltage (Kirchhoff's second law):
\[ V_{\text{total}} = V_{10\,\Omega} + V_{\text{parallel pair}} = 2.5 + 3.3 = 5.8 \text{ V} \]
This is approximately 6.0 V, consistent with the battery voltage. The small difference (0.2 V) can be attributed to contact resistance, ammeter resistance, or slight measurement uncertainties. [1]
(g) Both the 22 Ω and 33 Ω resistors are in parallel, so they have the same potential difference across them (3.3 V). Applying \( I = \frac{V}{R} \), a lower resistance gives a larger current. Since 22 Ω is less than 33 Ω, the 22 Ω resistor carries the larger current (0.15 A compared to 0.10 A). [1]
(h) Switch off the circuit before disconnecting and reconnecting the ammeter at a different position. This prevents damage to the ammeter and avoids accidental short circuits. [1]
(i) One source of error is the resistance of the ammeter itself, which adds to the circuit resistance and slightly reduces the current readings from their true values. Other valid sources include contact resistance at the connections, or the battery voltage dropping during the experiment due to internal resistance. [1]
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