Question

Three resistors of 4 ,6 and 12 are connected in parallel and the combination is connected in series with a 1.5 v battery of 1 internal resistance. The rate of joule heating in the 4 is:

Answer 1

Answer:

Calculating the Total Resistance of the 4, 6, 12 ohm parallel combination turns out to be 2 ohms.  

Adding that to the 2 ohms of the internal resistance of the battery, gives us a total load of 4 ohms.  

The current flow will then be 1 amp. 4v/4ohm=1 amp.  

The voltage across the parallel combination of 3 resistors will be 2 volt.  

The current through the 4 ohm resistor will be the voltage divided by the resistance, 2V/4ohm=1/2 Amp.  

The current through the 6 ohm resistor will be 2V/6ohm or 2/6 Amp.  

The current through the 12 ohm resistor will be 2V/12ohm or 2/12 amp or 1/6 amp.  

Adding the 3 currents together we get:

1/2 + 2/6 + 1/6 or 3/6 + 2/6 + 1/6 = 6/6 or 1 amp