Pls help with this question, I need the explanation/step by step solution.
Thanks in advance!
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well as the ammeter and the resistor are connected in series their resistances add up which happens to be 5.2 ohm. also the internal resistances of the battery is 2ohm so total resistance of the circuit is 5.2+2 which is 7.2 ohm.
(a) ammeter will read current which is V/R which is 1.8/7.2 which is 1/4 which is 0.25 A
(b) the potential difference across the terminals of the cell is V=IR which is 1/4*2 which is 0.5 A
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