|
Answer» the ammeter reading decreases the potential difference between E and F increases the potential difference between E and F stays the same bulb 3 LIGHTS up more brightly. Solution :Since FCD shorts bulb 3, the circuit I equivalently wires BC and EF in parallel. As the voltage across each BRANCH must be equal to the voltage of the battery, the voltage from E to F is unchanged by closing the switch. INITIALLY, there is voltage across the switch and after it is closed, the voltage is across the resistor. By closing the switch, the current in the circuit increases as effective resistance a cross battery decreases. (across battery effective resistance `=R_(1)/(1+(R_(1)//R_(2)) lt R_(1)` and the power from the battery must increase.
|