A series LR circuit has time constant τ. If resistance is doubled while inductance remains unchanged, how does the time required to reach 90% of final current change?
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A series LR circuit has time constant τ. If resistance is doubled while inductance remains unchanged, how does the time required to reach 90% of final current change?

G1 Physics · Accepted Answer

t90=−τln0.1 and τ=L/R. Doubling R halves τ and therefore halves t90.