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There are two bulbs in series in the circuit. The voltage at both ends of bulb L1 is 3V, the total resistance of the circuit is 10 ohm, and the resistance of bulb L1 is 5 ohm (1) Resistance of bulb L2 (2) current in circuit (3) voltage at both ends of bulb L2 (4) supply voltage
Because it is a series circuit, the total resistance R = R1 + R2, so the resistance of L2 is r-r1 = 5 Ω, the current in the series circuit is equal, I = 3V / 5 Ω = 0.6A, the voltage at both ends of bulb L2 is 0.6x5 Ω = 3V, and the power supply voltage is 3V + 3V = 6V
If a 1.5V battery is connected in parallel with a 9V battery and connected with a 1 ohm resistor, what is the current and voltage
If a 1.5V battery is connected in parallel with a 9V battery and a 1 ohm resistor is connected, what is the voltage at both ends of the resistor and the current and voltage of the branch
In general, two power supplies can not be directly connected in parallel! Because even the same power supply, due to temperature drift and voltage ripple and other reasons, the voltage can not be consistent, and a small voltage difference will lead to a great imbalance of power distribution! It may damage the power supply!
If connected in parallel by diodes, it can be used as redundant standby power supply!
The internal resistance of the battery is 0.2 ohm, the terminal voltage on the external circuit is 1.8V, and the current intensity in the circuit is 0.5A. Calculate the electromotive force and external resistance of the battery. The score is not much, but thank you
First of all, the electromotive force of the battery = internal resistance partial voltage + external circuit partial voltage
Internal resistance partial voltage = 0.5 * 0.2 = 0.1V
So the EMF of the battery is 0.1 + 1.8 = 1.9V
External resistance = external voltage / current = 1.8 / 0.5 = 3.6 Ω
Connect an unknown resistance R and a resistance with R1 = 20 ohm in series into the circuit, and the current flowing through R1 is 0.4A. Connect R and a resistance with R2 = 60 ohm in series into the same power supply, and the current flowing through R2 is 0.2A. Calculate the resistance value and power supply voltage of the resistance
If the power supply voltage is set to u, then according to the law of series circuit, there are:
U=0.4(R+20)
U=0.2(R+60)
That is, 0.4 (R + 20) = 0.2 (R + 60)
The solution is r = 20 Ω
Substituting into the first formula, u = 16V
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