When batteries of EMF E 1 and E 2 (E 1 > E 2 ) are connected in series opposing (positive terminals facing each other):

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When batteries of EMF E1 and E2 (E1 > E2) are connected in series opposing (positive terminals facing each other):

Answer: Net EMF = E 1 -E 2.

A Net EMF = E1+E2
B Net EMF = E1-E2
C Net EMF = 0
D Net EMF = E1 x E2

Correct answer: B. Net EMF = E1-E2

Explanation: Series opposing (aiding against each other): net EMF = E1 - E2 (EMFs subtract).

In series, the same current flows through both resistors; in parallel, the current splits between branches and the voltage across each resistor is the same.

Concept context

Ohm's law, resistors, Kirchhoff's laws, Wheatstone bridge, and cells.

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When batteries of EMF E<sub>1</sub> and E<sub>2</sub> (E<sub>1</sub> > E<sub>2</sub>) are connected in series opposing (positive terminals facing each other):

Concept context

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