When three 30
Ω
resistors are connected in parallel across a 9-volt source, the total current is

Question

When three 30 
Ω
resistors are connected in parallel across a 9-volt source, the total current is

BlackTom AI · Accepted Answer

When three 30 Ω resistors are connected in parallel across a 9 V source, the currents in the branches add up to give the total current.
Option A: 0.45 A. This would imply either each branch carries 0.15 A or an error in adding branch currents. Since each 30 Ω branch with 9 V across it draws I = V/R = 9/30 = 0.3 A, the total would be 3 × 0.3 A = 0.9 A, not 0.45 A.
Option B: 0.30 A. This value corresponds to the current through a single 30 Ω resistor at 9 V, not three parallel resistors. In parallel, currents sum, so the total must be larger than any one branch current.
Option C: 0.90 A. This matches the correct total current. Explanation: each branch draws I_branch = V/R = 9 V / 30 Ω = 0.3 A. With three identical resistors in parallel, total current I_total = 3 × 0.3 A = 0.9 A. Alternatively, the equivalent resistance of the three in parallel is R_eq = 30 Ω / 3 = 10 Ω, so I_total = V / R_eq = 9 V / 10 Ω = 0.9 A. This confirms 0.90 A as the total current.
Option D: 0.10 A. This is far too small for the configuration; it corresponds to perhaps misapplying Ohm’s law with a much larger resistance (e.g., thinking there is 90 Ω total or miscounting branches). The correct parallel combination yields a much smaller effective resistance and a correspondingly larger current than 0.10 A.

Physics 1 Hon - Bowles - p06/p07 P10.13/10.15 Electrostatics & Circuits MC Practice

When three 30 Ω resistors are connected in parallel across a 9-volt source, the total current is

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