Current Electricity: Speed-Solving Techniques (4)

easy • 2 min read
Tags Current Electricity

Question

In a circuit with battery EMF ε=12\varepsilon = 12 V, internal resistance r=1 Ωr = 1\,\Omega, and external resistors R1=4 ΩR_1 = 4\,\Omega and R2=6 ΩR_2 = 6\,\Omega in parallel, find the current drawn from the battery and the terminal voltage. Solve in under 60 seconds.

Solution — Step by Step

For two resistors in parallel, Req=R1R2R1+R2=4×64+6=2410=2.4 ΩR_{\text{eq}} = \dfrac{R_1 R_2}{R_1 + R_2} = \dfrac{4 \times 6}{4 + 6} = \dfrac{24}{10} = 2.4\,\Omega. The “product over sum” shortcut works only for two resistors in parallel — use it.

The internal resistance is in series with ReqR_{\text{eq}}: total Rtotal=r+Req=1+2.4=3.4 ΩR_{\text{total}} = r + R_{\text{eq}} = 1 + 2.4 = 3.4\,\Omega.

By Ohm’s law applied to the full circuit:

I=εRtotal=123.4≈3.53 AI = \frac{\varepsilon}{R_{\text{total}}} = \frac{12}{3.4} \approx 3.53 \text{ A}

Terminal voltage = EMF − IrIr = 12−3.53×1=8.4712 - 3.53 \times 1 = 8.47 V. Equivalently, terminal voltage = I×Req=3.53×2.4≈8.47I \times R_{\text{eq}} = 3.53 \times 2.4 \approx 8.47 V. Both give the same answer — pick whichever is faster.

Why This Works

The whole circuit collapses to a single loop once we combine the parallel resistors. Then it is just a 1-step Ohm’s law calculation. Internal resistance is in series with the external network, never in parallel — that is the rule that catches students.

Three speed shortcuts for current electricity:

  1. Two resistors in parallel: product over sum.
  2. Equal resistors in parallel: Req=R/nR_{\text{eq}} = R/n for nn identical resistors.
  3. Terminal voltage drop: V=ε−IrV = \varepsilon - Ir. Memorize this — it appears in 30% of NEET circuit questions.

Alternative Method

Apply Kirchhoff’s voltage law directly. Going around the loop: ε=I(r+Req)⇒I=ε/(r+Req)\varepsilon = I(r + R_{\text{eq}}) \Rightarrow I = \varepsilon / (r + R_{\text{eq}}). Same answer, but takes longer because you have to write the loop equation explicitly.

Students try to use the parallel formula for three or more resistors as Req=(R1R2R3)/(R1+R2+R3)R_{\text{eq}} = (R_1 R_2 R_3)/(R_1 + R_2 + R_3). That is wrong. Product-over-sum is a two-resistor shortcut only. For three or more, use 1Req=1R1+1R2+1R3\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}.

Final answer: I≈3.53I \approx 3.53 A, terminal voltage ≈8.47\approx 8.47 V.

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