✔ 最佳答案
┌──│ l──┐
│─A ─Ω──│
└──V───┘
if the light bulb is shorted, very large current will pass through the ammeter (since the shorted light bulb has no resistance), the ammeter will burn out. No reading is shown by ammeter. The voltmeter across the cell measures the cell emf (potential difference across the cell)
if the light bulb burns out, the circuit is open and no current pass through the ammeter. No reading is shown by ammeter too. The voltmeter again across the cell measures the cell emf.
┌──│l───┐
A │
├───Ω── ┤
└──V─── ┘
if the light bulb is shorted, very large current will pass through the ammeter too, the ammeter will burn out. No reading is shown by ammeter.
But at this time the voltmeter measures the p.d. across a wire, so its reading is zero.
if the light bulb burns out, in fact no current will be drawn as the resistance of voltmeter is infinitely large. if it is not an ideal voltmeter, extremely small current will be drawn but it can hardly be shown by the ammeter, it seems to have no resding.
