Explanations & Calculations
- Since the batteries are connected in series, their voltages add up to ("\\small 1.5V+1.5V=" ) "\\small3.0V" & their resistances are also connected in series.
- Moreover, these two resistances are also series connected with the external resistor giving the equivalent resistance of the circuit to be ("\\small 1\\Omega+1\\Omega+2\\Omega=") "\\small 4\\Omega".
- Now the circuit can be simplified to a "\\small 3.0V" battery connected to a "\\small 4\\Omega" resistor.
- Now simply using "\\small V=iR" the current flowing in the circuit can be calculated.
"\\qquad\\qquad\n\\begin{aligned}\n\\small i&=\\small \\frac{V}{R}=\\frac{3.0V}{4\\Omega}\\\\\n&=\\small\\bold{0.75A=750mA}\n\\end{aligned}"
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