Question

The emf ε and the internal resistance r of the battery, shown in the figure (32-E23), are 4.3 V and 1.0 Ω respectively. The external resistance R is 50 Ω. The resistances of the ammeter and voltmeter are 2.0 Ω and 200 Ω respectively. (a) Find the readings of the two 200 Ω respectively. (a) Find the readings of the two meters. (b) The switch is thrown to the other side. What will be the readings of the two meters now?
figure 32-E23

Answer 1

Answer:

(a) The 50 Ω and 200 Ω resistances are in parallel. Their equivalent resistance,

Reqv = (50×200)/(50+200)

= 40 Ω

This equivalent resistance and the 2 Ω and 1 Ω resistances are connected in series. The effective resistance of the circuit,

Reff = (40+2+1)

=43 Ω

In this case, the ammeter will read the total current of the circuit. The current through the ammeter, i = 4.3/43

= 0.1 A

This current will be distributed in the inverse ratio of resistance between the resistances 50 Ω and 200 Ω. The current through the voltmeter,

i' = 50/(50+200)i

⇒i' = 0.02 A

Reading of the voltmeter = 0.02 × 200 = 4 V

(b) The two branches AB and CD are in parallel. Their equivalent resistance,

Reqv = (52×200)/(52+200)

= 41.27 A

This equivalent resistance is in series with the 1 Ω resistance. The effective resistance of the circuit,

Reff = (41.27+1) Ω

= 42.27 Ω

The total current through the circuit,

i = 4.3/42.2.

= 0.1 A

In this case, the ammeter will read the current flowing through the 50 Ω resistance, which is i1, as shown. The currents in the two parallel branches will distribute in the inverse ratio of the resistances.

∴i1 = (200/200+52)i

⇒ i1 = 0.08 A

The current through the voltmeter = i - i1 = 0.02 A

The reading of the voltmeter = 0.02 × 200 = 4 V

Answer 2

Explanation:

(a) The resistances of 200 Ω and 50 Ω are in parallel. The equivalent resistance is given by,

Req = 20050 $\times$ 50 + 200 = 40 Ω

• The 2 Ω and 1 Ω resistances and this equivalent resistance are linked in series. In a circuit, the effective resistance, Reff = 40 + 1 + 2 = 43 Ω

• In this scenario, the circuit’s total current will be read by the ammeter. The current via the ammeter, I = 4.343 = 0.1 A

• In the inverse resistance ratio between the resistances 200 Ω and 50 Ω, this current will be spread. Via the voltmeter, the current is i’ = 200i + 5050 ⇒i’ = 0.02 A.

Voltmeter’s reading = 200 $\times$ 0.02 = 4 V

(b) AB and CD are the branches that are in parallel. Their equivalent resistance,

Reqv = 52 × 200 + 20052 = 41.27 A

• With the 1 Ω resistance, this equivalent resistance is in series. The circuit’s effective resistance, Reff = + 1 Ω + 41.27 = 42.27 Ω ,The total current, I = 0.1 A.

• In this scenario, the reading of the current flowing will be taken by the ammeter via the 50 Ω resistance, which is i1, as displayed. In the two parallel branches, the currents will allocate in the resistances’ inverse ratio.

Therefore, i1 = 52i + 200200 ⇒ ii = 0.08 A

The current passing via the voltmeter = 0.02 A.

The voltmeter reading = 200 $\times$ 0.02 = 4 V.