Two bulbs are connected in series to a 3 V battery.  The potential difference across one of the bulbs is 1.4 V.  Calculate the potential difference across the second bulb.1.4 V4.4 V1.5 V1.6 V2A resistor, a motor and a switch are connected in series to a 240 V power supply.  When the switch is closed, a current of 4 A flows through the resistor.  What can be said about the current through the motor?It is impossible to sayIt will be greater than 4 AIt will be equal to 4 AIt will be less than 4 A3A current of 0.8 A flows into a section of a circuit in which two bulbs are connected in parallel.  The current through the first bulb is 0.5 A.  Calculate the current through the second bulb.0.3 A1.3 A0.4 A1.6 A4Two 10 Ω resistors are connected in parallel.  Which of the following is the only possible total (equivalent) resistance of this combination of resistors?5 Ω100 Ω10 Ω20 Ω5Two resistors are connected in parallel to a 12 V battery.  The potential difference across one of the resistors is 12 V.  Calculate the potential difference across the other resistor.12 V1 V0 V144 V6A battery is connected in series to a bulb and a variable resistor.  Which of the following will NOT occur when the resistance of the variable resistor is decreased?The current in the circuit will increaseThe brightness of the bulb will increaseThe potential difference across the variable resistor will decreaseThe potential difference across the bulb will decrease

Sure, here are the answers to your questions:

1. The potential difference across the second bulb will be 1.6 V. This is because in a series circuit, the total voltage is the sum of the voltages across each component. So, 3V (total voltage) - 1.4V (voltage across first bulb) = 1.6V (voltage across second bulb).

2. The current through the motor will be equal to 4 A. In a series circuit, the current is the same through all components.

3. The current through the second bulb will be 0.3 A. In a parallel circuit, the total current is the sum of the currents through each component. So, 0.8A (total current) - 0.5A (current through first bulb) = 0.3A (current through second bulb).

4. The total (equivalent) resistance of two 10 Ω resistors connected in parallel is 5 Ω. The formula for total resistance (Rt) in a parallel circuit is 1/Rt = 1/R1 + 1/R2. So, 1/Rt = 1/10 + 1/10 = 2/10. Therefore, Rt = 10/2 = 5 Ω.

5. The potential difference across the other resistor will be 12 V. In a parallel circuit, the potential difference (voltage) is the same across all components.

6. The potential difference across the bulb will not decrease when the resistance of the variable resistor is decreased. In a series circuit, the potential difference across each component is determined by its resistance. If the resistance of the variable resistor decreases, the potential difference across it will also decrease. However, this will cause the potential difference across the bulb to increase, not decrease, because the total potential difference (provided by the battery) is constant.