Elec Class

learn electricity hands-on
0 1 / 20

Solved exercise · level 3 of 5

Cells in opposition

Skills: Analyse

Statement

Look at the polarity of the cells on the board: E1 = 9 V and E2 = 4.5 V are wired in opposition. Check the terminal directions, then compute the current in R1 = 10 Ω.

Parametric exercise: the values change with every draw; these are one example.

2 hints from Bastaya available in the lab
Solution

The EMFs subtract: I = |E1 − E2| / R1 = 450 mA. The weaker cell is crossed "backwards": it recharges (load).

Course sheet of the chapter : Kirchhoff's laws

This chapter is covered by the paths: General electricity, Measurement prep

Other exercises of the chapter

Simulated time00:00:00
Q = I·t-
W = P·t-

Circuit diagram CEI 60617

How does it work?

Each big hole is a circuit node: you push one leg of a component into it. The 4 small holes around it are connected to it: chips and leads plug in there.

  1. Place a component Drag it from the palette: it snaps onto the grid. R rotates it.
  2. Bridge neighbouring holes Pick the jumper chip (or C) and drag on the board: bridged holes become a single node.
  3. Draw a lead For long links and the power supply: from one big hole to another, or to a terminal.
  4. Close the loop The circuit is solved live (Ohm, Kirchhoff) and the diagram follows.

Shortcuts

Del
delete
R
rotate
C
chip brush
Esc
cancel
wheel
zoom
0
fit the board
drag background
pan

Confirmer

Renommer

Importer un montage

Export