Fig. 30.1 shows a circuit with a 6.0 V battery, two switches S₁ and S₂, and two lamps L₁ and L₂. S₁ is in series with L₁ and S₂ is in series with L₂. The tw...
Question 1 Report
Fig. 30.1 shows a circuit with a 6.0 V battery, two switches S₁ and S₂, and two lamps L₁ and L₂. S₁ is in series with L₁ and S₂ is in series with L₂. The two branches are connected in parallel.
L₁ is rated 6.0 V, 0.50 A. L₂ is rated 6.0 V, 0.30 A.
(a) Both switches are closed. State which lamps are on. [1]
(b) S₁ is opened and S₂ stays closed. State which lamp is on. [1]
(c) Calculate the power dissipated in L₁ when operating at its rated values. [2]
(d) Calculate the resistance of L₂ at its rated values. [2]
(e) State what happens to L₂ if L₁ blows. [1]
Answer Details
(a) With both switches closed, current flows through both parallel branches. Both L₁ and L₂ are on. [1]
(b) When S₁ is opened, it breaks the circuit through the L₁ branch, so L₁ goes off. S₂ remains closed, so current still flows through L₂. Only L₂ is on. [1]
(c) The power dissipated in L₁ at its rated values is found using \(P = IV\):
\[P = 0.50 \times 6.0\] [1]
\[P = 3.0 \text{ W}\] [1]
(d) The resistance of L₂ at its rated values is found using \(R = \frac{V}{I}\):
\[R = \frac{6.0}{0.30}\] [1]
\[R = 20 \;\Omega\] [1]
(e) If L₁ blows (its filament breaks), L₂ stays on and is unaffected. [1] Because the two lamps are on separate parallel branches, breaking one branch does not interrupt the current in the other. This is the key advantage of parallel wiring over series wiring.
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