(b)i. State two advantages of a lead-acid accumulator over a Leclanche cell.
ii. A parallel combination of 3\(\Omega\) and 4\(\Omega\) resistors is connected in series with a resistor of 4\(\Omega\) and a battery of negligible internal resistance. Calculate the effective resistance in the circuit.
(a) EMF / internal-resistance experiment (outline). With standard resistor \(S\) in series and box resistance \(R\), the current is \(I = \dfrac{E}{R + S + r}\). Rearranging,
\[ R = E\left(\frac{1}{I}\right) - (S + r). \]
So a graph of \(R\) (vertical) against \(I^{-1}\) (horizontal) is a straight line of slope \(s = E\) (the e.m.f. of the accumulator) and vertical-axis intercept \(C = -(S + r)\), from which the internal resistance is found.
Precautions: close the key only while taking a reading to avoid running down the cell; read the ammeter at eye level to avoid parallax and ensure clean, tight connections.
(b)(i) Two advantages of a lead-acid accumulator over a Leclanche cell: it can be recharged and reused (the Leclanche cell cannot), and it has a much lower internal resistance so it can deliver a large current.
(b)(ii) Parallel combination of \(3\ \Omega\) and \(4\ \Omega\):
\[ R_p = \frac{3\times4}{3+4} = \frac{12}{7} = 1.71\ \Omega. \]
In series with \(4\ \Omega\):
\[ R_{eff} = 1.71 + 4 = 5.71\ \Omega. \]