What is the current through the battery when the switch is open

Consider the circuit shown in figure a Find the current through the battery a long time after the switch S is closed b suppose the switch is opened at t 0. Requivalent 1 1 R1 1 R2 1 R3.

We are asked how does the current going through r1 so this resistor when the switch is open this switch compared to the current through r1 when the switch is closed pause this video and see if you can figure that out alright so lets just think about the two scenarios so we could view the current as this right over here this current that we care about we could either measure it there.

What is the current through the battery when the switch is open. What is the current through the battery when the switch is open. What is the current through the battery when the switch is closed. When the switch is open the current finds its way from the battery through the resistances at branch AC then through the resistances GH and EF and finally through the resistances at branch DB.

I repeat again the open switch is like it doesnt exist. What is the current through the battery immediately after the switch is closed. The switch is reopened at t 3ms.

RC τ time constant charge is. The switch in the figure has been open for a long time. It is closed at t 0s.

A What is the current through the battery immediately after the switch is closed. B What is the current through the battery after the switch has been closed a long time. Both switches are initially open and the capacitor is uncharged.

What is the current through the battery after switch 1 has been closed a long time. 1 I b 0 2 I b E3R 2R I b - 3 I b E2R 4 I b ER ε C R–S 2 S 1 Long time current through capacitor is zero I b0 because the battery and capacitor are in series. Lecture 7 Slide 8.

Answer 1 of 5. The following circuits shows the voltage across open and closed switches in a 12V circuit. These are simulation results but we can use some basic scientific principles to understand and predict the voltage across an open switch.

The battery voltage is V 12 V. The positive terminal of the battery is indicated with a sign. 1 The switch has been open for a long time when at time t 0 the switch is closed.

What is I 4 0 the magnitude of the current through the resistor R 4 just after the switch is closed. Rseries 20Ω. R1 R2 R3 8Ω.

V battery 20Volts. Let Requivalent be the equivalent resistance of the 3 resistors in parallel. Requivalent 1 1 R1 1 R2 1 R3.

Requivalent 1 1 8Ω 1 8. Ω 1 8Ω. The direction of current is from positive to negative end though in reality it is the opposite link The longer end of the battery is the positive end and the shorter end of the battery is the negative end.

The positive end of the ammeter is connected to the positive end of the battery while the negative end of the ammeter is. Throw switch S is open as shown in the figure the current in the battery is 100 mA. When the switch is closed in position 1 the current in the battery is 120 mA.

The switch in the figure below is initially open and the capacitor is uncharged. 600 V R 100 Ω and C 720 µF what current flows through the battery immediately after the switch is closed. A Because the switch is closed the battery gets connected across the LR circuit.

The current in the LR circuit after t seconds after connecting the battery is given by. I i 0 1 e t τ Here i 0 Steady state current. τ Time constant LR After a long time t.

Now Current in the inductor i i 0 1 e 0 0. For the circuit shown in the figure V 60 V C 20 µF R 010 MΩ and the battery is ideal. Initially the switch S is open and the capacitor is uncharged.

The switch is then closed at time t 000 s. What is the charge on the capacitor 80 s after closing the switch. The current through the battery 1 2 I must go through each of the 2 Ω-resistors connected at node A because V A V C 1 2 I2 Ω V A V D.

At node B the 2 Ω-resistor inputs twice the current of the 4 Ω -resistor or 2 3 I and 1 3 I respectively because V C V B 2 3 I2 Ω 1 3 I4 Ω V D V B. Therefore 1 6 I must go through the ammeter from D to. We are asked how does the current going through r1 so this resistor when the switch is open this switch compared to the current through r1 when the switch is closed pause this video and see if you can figure that out alright so lets just think about the two scenarios so we could view the current as this right over here this current that we care about we could either measure it there.

Metal ring Iron core Switch Coil Battery Step With the switch open what is the magnetic flux through the ring. Mark the direction of the current in the coil after the switch is closed Step 3 Draw magnetic field lines through the coil and the RING. Do you apply Right-Hand-Rule 1 or 2.

6 After being closed a long time switch 1 is opened and switch 2 is closed. What is the current through the right resistor immediately after the switch 2 is closed. E a IR 0 b IRE3 R c IRE2 R d IRER Now the battery and the resistor 2R are disconnected from the circuit so we have a different circuit.

Since C is fully charged. When the switch in the circuit presented by the diagram to the right is open the voltmeter reading is referred to as. What is the current through the battery.

Consider the circuit shown in figure a Find the current through the battery a long time after the switch S is closed b suppose the switch is opened at t 0. What is the time constant of the decay circuit. C Find the current through the inductor after one time constant.

Consider the circuit shown in figure a Find the current through the battery a long time after the switch S is closed b suppose the switch is opened at t 0. What is the time constant of the decay circuit. C Find the current through the inductor after one time constant.

Since there is 9Ω in the circuit and the voltage source is 15V the current going through the circuit is I 15V9Ω 017A. We now use a revised version of ohms law VIR to calculate the voltages across the battery and the resistor. Since the battery has an internal resistance of 1Ω it drops V 017A 1Ω 017V.

Consider the circuit below. The battery has an emf of ε 3000 V ε 3000 V and an internal resistance of r 100 Ω. R 100 Ω.

A Find the equivalent resistance of the circuit and the current out of the battery. B Find the current through each resistor. C Find the potential drop across each resistor.

D Find the power dissipated. Switch is closed bulbs A and B are in series. Bulb C is not part of the circuit.

A and B are identical resistors in series so their equivalent resistance is 2 R and the current from the battery is This is the initial current in bulbs A and B so they are equally bright. 2015 Pearson Education Inc.