Three identical cells of emf 2 V and internal resistances `0.20 Omega` are connected in series. The combination is further connected to an external resistor of `6 Omega` . Calculate the current through `6 Omega` resistor.

Explanation

To solve the problem of finding the current through a 6 Ω resistor connected in series with three identical cells, each with an emf of 2 V and an internal resistance of 0.20 Ω, we can follow these steps:

### Step 1: Calculate the Total EMF
The total electromotive force (emf) from the three cells connected in series can be calculated as follows:
\[ \text{Total EMF} (V_{\text{net}}) = \text{EMF}_1 + \text{EMF}_2 + \text{EMF}_3 = 2\, \text{V} + 2\, \text{V} + 2\, \text{V} = 6\, \text{V} \]

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